Simplify the expression by using a double-angle formula. 2 cos^2 4 theta -1

Answers

Answer 1
You are given 2cos² (4θ - 1). You are asked to simplify the expression by using a double-angle formula. 

Apply the double angle of the cosine function
cos2θ = 2cos² θ - 1

Rearranging the expression and then substituting them
2cos² θ = cos2θ + 1
2cos² (4θ - 1) = cos 2(4θ - 1) + 1
cos2(4θ - 1) + 1


Related Questions

A pile of dirt is cone shaped and it has a height of 10 feet and a diameter of 24 feet. Find the volume. * The answer is NOT 1570.96 or 1570.2*

Answers

You'll need to specify the accuracy to which you want your answer.  For example:  "Find the volume of this cone to 3 decimal place accuracy."

The formula for the volume of a cone is  V = (1/3) (base) (height), where "base" represents the area of the base.

Here the diameter of the base is 24 feet, so the radius of the base is 12 feet.
Thus, the area of the base is  (12 feet)^2 times pi:     A = 144 pi ft^2.

Multiply this area by the height of the cone, which is 10 feet:

V = 1440 pi ft^3

By calculator, this volume is  4523.893421 cubic feet.

If you want this volume to the nearest cubic foot, it'd be 4524 cubic ft.
If you want this vol. to the nearest 100th cubic ft., it'd be 4523.89 cu. ft.

William invested $6000 in an account that earns 5.5% interest, compounded annually. The formula for compound interest is A(t) = P(1 + i)t.

How much did William have in the account after 6 years? (APEX)

Answers

[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+r\right)^{t} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$6000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\to &0.055\\ t=years\to &6 \end{cases} \\\\\\ A=6000(1+0.055)^6\implies A=(1.055)^6[/tex]

Answer:

William have $8273.057 in the account after 6 years.

Step-by-step explanation:

The given formula is [tex]A(t)=P(1+i)^t[/tex]

We have,

P = $6000

r = 5.5% = 0.055

t = 6

A =?

Substituting these values in the above formula to find A

[tex]A(t)=6000(1+0.055)^6\\\\A(t)=8273.057[/tex]

Therefore, William have $8273.057 in the account after 6 years.

Your science quiz had 17 questions and you answered 13 of the questions correctly. What is your present score?

Answers

dived 13 by 17

13/17 = 0.7647

  = 76.47%

 if you need to round the number it would be 76%, which is a 76 grade

When would it be useful to sort data in descending order? why?

Answers

when seeing who has the highest score on a test

Events A and B are mutually exclusive with P(C) = 0.3 and P(B) = 0.2. Then P(Bc) =

Answers

If 2 events are mutually exclusive then probability they both occur is 0.
Final answer:

The probability of the complement of event B, denoted as Bc, is 0.8.

Explanation:

To find the probability of the complement of an event B, denoted as Bc, we can use the formula: P(Bc) = 1 - P(B). Given that events A and B are mutually exclusive, P(B) = 0.2. Therefore, P(Bc) = 1 - 0.2 = 0.8.

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If f(x) is a nth degree polynomial then F^(n+1)(x)=0. True or false and why

Answers

Final Answer:

The given statement “If f(x) is a nth degree polynomial then [tex]F^{(n+1)(x)=0[/tex]. True or false and why”  is false because the (n+1)st derivative being zero is contingent on the specific value of the leading coefficient in the polynomial. It is not a general rule for all nth degree polynomials.

Explanation:

The statement [tex]\(F^{{(n+1)}(x) = 0\)[/tex] is not universally true for all nth degree polynomials (f(x)). To understand why, consider a general nth degree polynomial [tex]\(f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\)[/tex], where [tex]\(a_n\)[/tex] is the leading coefficient and [tex]\(a_n \neq 0\)[/tex].

The nth derivative of (f(x)) can be expressed as [tex]\(f^{(n)}(x) = n! \cdot a_n\).[/tex]Now, the (n+1)st derivative, [tex]\(f^{(n+1)}(x)\)[/tex], will be zero if and only if the leading coefficient [tex]\(a_n = 0\)[/tex]. However, this condition is not satisfied in general, as [tex]\(a_n\)[/tex] is assumed to be nonzero for a nontrivial polynomial. Therefore, the (n+1)st derivative is not guaranteed to be zero for all nth degree polynomials.

In mathematical terms,[tex]\(F^{(n+1)}(x)\)[/tex] equals zero if and only if the leading coefficient of (f(x)) is zero, but this is not a universal characteristic of nth degree polynomials. Consequently, the statement is false, and the (n+1)st derivative may not be zero for all x in the domain of the polynomial.

Suppose you value a special watch at $100. you purchase it for $75. on your way home from class one day, you lose the watch. the store is still selling the same watch, but the price has risen to $85. assume that losing the watch has not altered how you value it. what should you do?

Answers

I will search all the way back to class and if i couldn't find it i will buy a new one.

coz $10 is not a very big amount.

What is 243.875 rounded to the nearest tenth,hundreth,ten,and hundred?

Answers

tenth: 243.9
hundreth: 243.88
ten: 240
hundred:200

hope this helps

A carpenter is framing a window with wood trim where the length of the window is 9 1/3 feet. If the width of the window is 6 3/4 feet, how many feet of the wood is needed to frame the window?

Answers

This is not much of a question, because it doesn't take into account the width of the wood trim.  

So you are asked to find the perimeter of the window, ignoring the width of the trim.

P = Perimeter = 2L + 2W.

Here, P= 2(9 1/3 feet) + 2(6 3/4 feet) = 2(28/3 feet) + 2(27/4 feet)

So P = 2[28/3 + 27/4] feet.  LCD is 3*4 = 12.

Thus, P = 2 [ 28/3 + 27/4 ] feet.  Can you finish?  Add together the fractions 28/3 and 27/4.

If you toss six fair coins, in how many ways can you obtain at least two heads?

Answers

1/3 that is the correct answer

starting at home, luis traveled uphill to the hardware store for 30 minutes st just 8mph. he then traveled back home along the same path downhill at a speed of 24 mph. what is his average speed for the entire trip from home to the hardware store and back?

Answers

1.
The main formula we use is :

Distance traveled = average Speed * Time,

in short:    D = S * T

2.
"luis traveled uphill to the hardware store for 30 minutes at speed 8mph."

we have the speed and the time = 30 min = 1/2 h

so we can find distance:

D=ST=8 (mi/h) * 1/2 (h)  =  4 mi

3.
Now we have D, and the speed 24 mph, so we can find the time it takes Luis to travel back home:

D=S*T

T= D/S = 4 (mi) / 24 (mi/h) =1/6 h

4.

The total distance is D+D = 4 +4 = 8 (mi)

the total time is 1/2 +1/6 = 3/6 + 1/6 = 4/6 = 2/3 (h)

So we can find the average speed of the entire trip is:

S=8 / (2/3)= 8* (3/2)=4*3=12 (mph)


Answer : 12 mph 


find the binomial coefficient: 2012/2011

Answers

 ²⁰¹²C₂₀₁₁ = (2012)! / [(2011)! (2012-2011)!]

²⁰¹²C₂₀₁₁ = (2012)! / [(2011)! (1)!]

Simplify 2012! / (2011)! = 2012

²⁰¹²C₂₀₁₁ = (2012)! / (1)!  = 2012
Final answer:

The binomial coefficient '2012 choose 2011' is calculated using the formula C(n,k) = n! / [(n-k)! * k!]. After substituting the respective values into the formula, we find that the binomial coefficient of '2012 choose 2011' is 2012.

Explanation:

The binomial coefficient, often referred to in Mathematics, is generally expressed as 'n choose k' and calculated using the formula: C(n,k) = n! / [(n-k)! * k!]. In this formula, the '!' denotes factorial which means the product of an integer and all the integers below it.

However, the student's question seems to be asking for the binomial coefficient of '2012 choose 2011', which is misinterpreted as a fraction instead.

To calculate it correctly, we would apply the formula mentioned before: C(2012,2011) = 2012! / [(2012-2011)! * 2011!]. Because 2012-2011 equals 1, this simplifies our calculation. The factorial of 1 is 1 itself. Thus, C(2012,2011) = 2012!/ (1! * 2011!), which simplifies to be 2012. So the binomial coefficient of '2012 choose 2011' is 2012.

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If f(x) = x - 5, then match each of the following.

1. f(-1) -3
2. f(0) -6
3. f(1) -5
4. f(2) 0
5. f(5) 3
6. f(8) -4

Answers

f(-1) = -6
f(0) = -5
f(1) = -4
f(2) = -3
f(5) = 0
f(8) = 3
hello : 
1. f(-1) -3 = -1-5-3 = -9
2. f(0) -6   = 0-5-6 = -11
3. f(1) -5 = 1-5-5 =-9
4. f(2) 0 ??? +0 or : ×0
5. f(5) 3 ....
6. f(8) -4....
f(-1)-3 = f(1)-5

If marc ABC = 184°, what is m∠ABC?

Answers

the angle is 1/2 of the arc

184/2 = 92 degrees

It would be half of the intercepted arc which is

360-184 = 176

176/2 = 88 degrees

a dime is flipped 3 times. What is the probability that TAILS occurred all 3 times?

Answers

1/2*1/2*1/2=1/8

Therefore, the probability of tails being occurred 3 times is 1/8.

Hope this helps!
For three tosses of the coin all the possible outcomes are:
.
H-H-H
T-H-H
H-T-H
H-H-T
T-H-T
T-T-H
H-T-T
T-T-T

These eight possible outcomes are the sample space.
To find the probability of tossing tails tree times look down the sample space list and find any outcome that has exactly two H's.
The possibilities are only T-T-T.
This means that one of the eight possible outcomes contain exactly three tails. Therefore,
the probability of throwing exactly two heads in three tosses of the coin is 1 out of 8,

Every evening Jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she had 38 coins all of them dimes and quarters. When she added them up she had a total of $6.95

Answers

d = dimes

q = quarters

d+q = 38 coins

q=38-d

0.25q + 0.10d=6.95

0.25(38-d)+0.10d=6.95

9.5-0.25d+0.10d=6.95

-015d=-2.55

d=-2.55/-0.15 = 17

q=38-17 =21

21*0.25 =5.25

17*0.10 = 1.70

5.25+1.70 = 6.95

 she had 21 quarters and 17 dimes

Answer:

She had 17 dims and 21 quarters to make $6.95.

Step-by-step explanation:

Given : Every evening Jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she had 38 coins all of them dimes and quarters.

To find : When she added them up she had a total of $6.95?

Solution :

Let d be the dims and q be the quarters.

She had 38 coins.

i.e. [tex]d+q=38[/tex] ......(1)

The value of d is 0.10 and q is 0.25.

The total she had of $6.95

i.e. [tex]0.10d+0.25q=6.95[/tex]  .......(2)

Solving (1) and (2),

Substitute d from (1) into (2)

[tex]0.10(38-q)+0.25q=6.95[/tex]

[tex]0.10\times 38-0.10q+0.25q=6.95[/tex]

[tex]3.8+0.15q=6.95[/tex]

[tex]0.15q=6.95-3.8[/tex]

[tex]0.15q=3.15[/tex]

[tex]q=\frac{3.15}{0.15}[/tex]

[tex]q=21[/tex]

Substitute in equation (1),

[tex]d+21=38[/tex]

[tex]d=38-21[/tex]

[tex]d=17[/tex]

Therefore, She had 17 dims and 21 quarters to make $6.95.

An arithmetic sequence is represented in the following table. Enter the missing term of sequence

Answers

In mathematics, numbered sequential patterns are distinguished as progressions. There are three types of progression: arithmetic, geometric and harmonic. Let's focus on the arithmetic progression.

The pattern in the arithmetic progression is the common difference, You will find that when you subtract two consecutive terms of the sequence, you would get a common difference. Let's investigate further:

28-44 = -16
12-28 = -16
-4-12 = -16

Thus, the common difference is -16. To know the last term, just simply add -16 to the very last known term. In this case, -4+-16 = -20. The answer is -20.

Answer:

The required 18th term of the given sequence will be -160

Step-by-step explanation:

The A.P. is given to be : 44, 28, 12, -4, ....

First term, a = 44

Common Difference, d = 28 - 44

                                       = -12

We need to find the 18th term of the sequence.

[tex]a_n=a+(n-1)\times d\\\\\implies a_{18}=44+(18-1)\times -12\\\\\implies a_{18}=44+ 17 \times -12\\\\\implies a_{18}=44-201\\\\\implies a_{18}=-160[/tex]

Hence, The required 18th term of the given sequence will be -160

1. Suppose y varies directly with x. Write a direct variation equation that relates x and y. Then find the value of y when x= 12.

y = -10 when x = 2

2. Graph the direct variation equation:

y=2x

Answers

[tex]\bf \qquad \qquad \textit{direct proportional variation}\\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array}\\\\ -------------------------------\\\\[/tex]

[tex]\bf \textit{\underline{y} varies directly with \underline{x}}\implies y=kx \\\\\\ \textit{we also know that } \begin{cases} y=-10\\ x=2 \end{cases}\implies -10=k2\implies \cfrac{-10}{2}=k \\\\\\ -5=k\qquad thus\qquad \boxed{y=-5x}\\\\ -------------------------------\\\\ \textit{what's \underline{y} when \underline{x} is 12?}\qquad y=-5(12)[/tex]

The marching band is selling cases of fruit for $13 per case. (a) Write an algebraic expression for the cost of f cases of fruit. (b) Evaluate the expression for 250 cases.

Answers

(a) Each case of fruit costs $13
∴ $13 × f cases
= 13f

(b) Since f = 250
= 13(250)
= $3250

An algebraic expression for the cost of f cases of fruit will be (a) z = 13f and the cost for 250 cases will be (b) $3250.

How to form an equation?

Determine the known quantities and designate the unknown quantity as a variable while trying to set up or construct a linear equation to fit a real-world application.

In other words, an equation is a set of variables that are constrained through a situation or case.

Let's say the cost of the fruit is z

Given,

The marching band is selling cases of fruit for $13 per case.

So for f cases of fruit

z = 13f

And cost of 250 cases = z = 13(150) = $3250

Hence, the algebraic expression for the cost of f cases of fruit will be z = 13f and the cost for 250 cases will be $3250.

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A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into the tank at a rate of 6 gal/min. the well-mixed solution is pumped out at the same rate. find the number a(t) of pounds of salt in the tank at time t.

Answers

If [tex]A(t)[/tex] is the amount of salt in the tank at time [tex]t[/tex], then the rate at which the amount of salt in the tank changes is given by

[tex]\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}[/tex]
[tex]\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}[/tex]

Let's drop the units for now. We have

[tex]\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24[/tex]
[tex]e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}[/tex]
[tex]e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt[/tex]
[tex]e^{t/100}A(t)=2400e^{t/100}+C[/tex]
[tex]A(t)=2400+Ce^{-t/100}[/tex]

We're given that the water is pure at the start, so [tex]A(0)=0[/tex], giving

[tex]A(0)=0=2400+Ce^{-0/100}\implies C=-2400[/tex]

So the amount of salt in the tank (in lbs) at time [tex]t[/tex] is

[tex]A(t)=2400\left(1-e^{-t/100}\right)[/tex]
Final answer:

To find the amount of salt in the tank at a given time, one can use the equation a(t) = Q - Qe^(-rt). In this case, Q (the quantity of salt at a steady state) equals the pump rate multiplied by the salt concentration (24lb/min), and r (the rate of inflow and outflow of the solution) is the rate at which water is pumped out divided by the volume of the tank (1/100 per min). Substituting these values into the equation gives the salt content at any given time.

Explanation:

The quantity of salt in the tank at any given time can be determined by the equation of the form a(t) = Q - Qe^(-rt), in which Q is the quantity of salt that would be in the tank at a steady state (i.e., if enough time had passed that the quantity of salt in the tank stopped changing), r is the rate of inflow and outflow of the solution, and t is the time at which you're trying to determine the number of pounds of salt in the tank.

In this case, Q = rate of inflow x concentration of the inflow, which is 6 gal/min x 4 lb/gal = 24 lb/min. This amount is reached after a sufficient amount of time has passed and the tank has reached a steady state.

The rate, r, is the rate at which the water is pumped out of the tank. In this situation, that's 6 gallons per minute. Since there are 600 gallons of water in the tank at the start, r = 6 gal/min ÷ 600 gallons = 1/100 min^-1.

Therefore, the number of pounds of salt in the tank at any time t is a(t) = Q - Qe^(-rt) = 24 lb/min - 24 lb/min * e^[-(1/100 min^-1)*t].

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Suppose that 19 inches of wire costs 95 cents. At the same rate, how much (in cents) will 37 inches of wire cost?

Answers

0.95/19 = 0.05 cents per inch

37*0.05 = 1.85

it will cost $1.85

Which of the following expressions represents a function?

x2 + y2 = 9
{(4, 2), (4, –2), (9, 3), (9, –3)}
x = 4
2x + y = 5

Answers

A. Not a function because it's a circle that doesn't pass the vertical line test

B. Not a function because we have x = 4 repeated more than once. Same for x = 9 as well.

C. Not a function. Any vertical line fails the vertical line test. Any vertical line is therefore not a function.

D. This is a function since it passes the vertical line test.

------------------------------------------------------

In summary, the final answer is choice D

D is the correct answer the other person who answered explains why.

The sales at a particular bookstore grew from $2090 million in 2000 to $3849 million in 2005. Find an exponential function to model the sales as a function of years since 2000. Give your answer using the form B=Boat

Answers

[tex]\bf \qquad \textit{Amount for Exponential Growth}\\\\ A=I(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &2090\\ I=\textit{initial amount}\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &0\\ \end{cases} \\\\\\ 2090=I(1+r)^0\implies 2090=I\\\\ -------------------------------\\\\[/tex]

[tex]\bf \qquad \textit{Amount for Exponential Growth}\\\\ A=I(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &3849\\ I=\textit{initial amount}\to &2090\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &\stackrel{2000-2005}{5}\\ \end{cases}[/tex]

[tex]\bf 3849=2090(1+r)^5\implies \cfrac{3849}{2090}=(1+r)^5\implies \sqrt[5]{\cfrac{3849}{2090}}=1+r \\\\\\ \sqrt[5]{\cfrac{3849}{2090}}-1=r\implies 0.129900823\approx r\implies 0.13\approx r\\\\ -------------------------------\\\\ A=2090(1+0.13)^t\implies \boxed{A=2090(1.13)^t}[/tex]

Picture Perfect Physician’s has 5 employees. FICA Social Security taxes are 6.2% of the first $118,500 paid to each employee, and FICA Medicare taxes are 1.45% of gross pay. FUTA taxes are 0.6% and SUTA taxes are 5.4% of the first $7,000 paid to each employee. Cumulative pay for the current year for each of its employees are as follows. Compute the amounts in the table for each employee and then total the numerical columns.
Employee
Cumulative Pay
Pay subject to FICA- S.S.
6.2% (First $118,000)
Pay subject to FICA-Medicare 1.45%
Pay subject to FUTA Taxes 0.6% Pay subject to SUTA Taxes 5.4% (First $7,000)

Mary $6,800
Type answer here
Type answer here
Type answer here
Type answer here

Zoe $10,500
Type answer here
Type answer here
Type answer here
Type answer here

Greg $8,400
Type answer here
Type answer here
Type answer here
Type answer here

Ann $66,000
Type answer here
Type answer here
Type answer here
Type answer here

Tom $4,700
Type answer here
Type answer here
Type answer here
Type answer here

Totals
Type answer here
Type answer here
Type answer here
Type answer here


Answers

Employee                                 Mary      Zoe         Greg         Ann           Tom

Cumulative Pay                       $6,800   $10,500  $8,400    $66,000   $4,700

Pay subject to FICA S.S.         $421.60  $651.00  $520.80 $4092.00 $291.40
6.2%, (First $118,000)

Pay subject to FICA Medicare $98.60 $152.25    $121.80    $957.00    $68.15
1.45% of gross

Pay subject to FUTA Taxes      $40.80  $63.00     $50.40    $396.00  $28.20
0.6%

Pay subject to SUTA Taxes   $367.20  $567.00  $453.60  $3564.00 $253.80
5.4% (First $7000)

Totals                                     $928.20 $1,433.25 $1,146.60 $9,009.00 $641.55

Answer:

Employee                                 Mary      Zoe         Greg         Ann           Tom

Cumulative Pay                       $6,800   $10,500  $8,400    $66,000   $4,700

Pay subject to FICA S.S.         $421.60  $651.00  $520.80 $4092.00 $291.40

6.2%, (First $118,000)

Pay subject to FICA Medicare $98.60 $152.25    $121.80    $957.00    $68.15

1.45% of gross

Pay subject to FUTA Taxes      $40.80  $63.00     $50.40    $396.00  $28.20

0.6%

Pay subject to SUTA Taxes   $367.20  $567.00  $453.60  $3564.00 $253.80

5.4% (First $7000)

Totals                                     $928.20 $1,433.25 $1,146.60 $9,009.00 $641.55

Step-by-step explanation:

Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 35 liters per minute. There are 700 liters in the pond to start. Let W represent the total amount of water in the pond (in liters), and let T represent the total number of minutes that water has been added. Write an equation relating W to T . Then use this equation to find the total amount of water after 19 minutes.

Answers

W=700+35T. Then replace T with 19 which will be W=700+35(19) which equals to 1365 total amount of water in liters.

An equation relating W to T is,

W = 700 + 35T

And, Total amount of water after 19 minutes is, 1365 liters

We have to given that,

Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 35 liters per minute.

There are 700 liters in the pond to start.

Now, Let W represent the total amount of water in the pond (in liters), and let T represent the total number of minutes that water has been added.

Hence, an equation relating W to T is,

W = 700 + 35T

So, For T = 19 minutes,

W = 700 + 35  x 19

W = 700 + 665

W = 1365

Therefore, Total amount of water after 19 minutes is, 1365 liters

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A chemist has three different acid solutions. The first acid solution contains
25%
acid, the second contains
40%
and the third contains
60%
. He wants to use all three solutions to obtain a mixture of
60
liters containing
45%
acid, using
3
times as much of the
60%
solution as the
40%
solution. How many liters of each solution should be used?

Answers

Let t, f, and s be the amount of 25%, 40%, and 60% solutions used.

We are told s=3f.  The general equation is:

100(0.25t+0.4f+0.6s)/(t+s+f)=45, and using s=3f  we have:

(0.25t+0.4f+1.8f)/(t+4f)=0.45

(0.25t+2.2f)/(t+4f)=0.45

0.25t+2.2f=0.45t+1.8f  

0.25t+0.4f=0.45t

0.4f=0.2t

t=2f, remembering that s=3f, will allow us to solve for f

t+s+f=60, using t and s from above

2f+3f+f=60

6f=60

f=10 L then

s=3f=30 L  and t=2f=20L

So 20L of 25%, 10L of 40%, and 30L of 60% acid solutions need to be mixed to make 60L of 45% acid solution.

which is larger 2/3" x 3-7/16" or 0.6"L x 3.43"W?

Answers

2/3 x (3 - 7/16) = (2/3) * (3+7/16)
       = (2/3)*(55/16)
       = 55/24
       = 2.2917

0.6 x 3.43
      = 2.058

The first answer is greater than the second.

Answer:  2/3 x 3-7/16 is larger.

Write an algebraic equation for the following problem and then solve it.
The population of a country in 2015 was estimated to be 321.6321.6 million people. This was an increase of 22​% from the population in 1990. What was the population of the country in​ 1990?

Answers

let's say is the population in 1990 is "x".

well, in 2015, 25 years later, it ballooned to 321,6321.6, and we know that's 22% or 22/100 ( 0.22 ), more than 25 years ago.

if the amount on 1990 was "x", then 22% of "x" is just (22/100) * x, which is 0.22x.

so, whatever "x" may be, the sum of those two is 321,6321.6, thus

[tex]\bf 321,6321.6 = x + 0.22x\implies 321,6321.6 = 1.22x\\\\\\ \cfrac{321,6321.6}{1.22}=x[/tex]

factoring
x^2-4x-21=0

Answers

Find two numbers that add to -4 and that multiply to -21.

After playing around with some numbers, I got -7 and 3.

So, the factored form is (x - 7)(x + 3).

A jeep and BMW enter a highway running east-west at the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the BMW did, and traveled 7 mph slower than the BMW. After 2 hours from the time the BMW entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control.

Answers

recall you d = rt, distance = rate * time

when the BMW has been running for 2hrs, the Jeep has been running for 2.5hrs, because it started off the exit 30minutes(0.5hr) before the BMW.

if the rate of the BMW is say "r", then the rate of the Jeep is "r - 7", slower by 7mph.

now, we know after the Jeep was running for 2.5hrs and the BMW was running for 2hrs, they were both apart by 306.5miles, so, say if the Jeep travelled a distance of "d", then the BMW travelled the slack from the 306.5, or "306.5 - d".

[tex]\bf \begin{array}{lccclll} &distance&rate& \begin{array}{cllll} time\\ (hours)\\ \end{array}\\ &-----&-----&-----\\ Jeep&d&r-7&2.5\\ BMW&306.5-d&r&2 \end{array} \\\\\\ \begin{cases} \boxed{d}=(r-7)2.5\\ 306.5-d=2r\\ ----------\\ 306.5-\boxed{(r-7)2.5}=2r \end{cases}[/tex]

solve for "r" to get the rate of the BMW.

what about the Jeep's? well, the Jeep is just r - 7.

Answer:

3x+27+3.5x=397.5

6.5x=370.5

x=57 mph for jeep and travels 199.5 miles in 3.5 hours

x+9=66 mph for B and travels 198 miles in 3 hours

Step-by-step explanation:

speed of jeep=x, time is     >>>>>    t+0.5 in hours or 3.5 hours here

speed of B=x+9, time is  >>>>>>>>>>>> 3 hours

distance is >>>>>>>>>    speed time

Hope it Helps.

Answer on: june 11, 2021

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