Sue uses the expression ( 8 x 3,000) + ( 8 x 200) + (8 x 9) to help solve a multiplucation problem what is Sue's multiplucation problem?

The correct answer is C.

Jonathan’s answer is reasonable because 2 times 8 is 16, and 16 is close to 15.77.

Hope this helps.

Answer:

Hence, AB=12.

Step-by-step explanation:

We are given that the perpendicular bisector of side AB of ∆ABC intersects side BC at point D.

this means that side AE=BE.

Also we could clear;ly observe that

ΔBED≅ΔAED

( since AE=BE, side ED common, ∠BED=∠AED

so by SAS congruency the two triangles are congruent)

Now we are given that:

the perimeter of ∆ABC is 12 cm larger than the perimeter of ∆ACD.

i.e. AB+AC+BC=AC+AD+CD+12

AB+BC=AD+CD+12

as AD=BD

this means that AD+CD=BD+CD=BC

AB+BC=BC+12

AB=12

Hence AB=12

Answer:

The required length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

Step-by-step explanation:

Given: The perpendicular bisector of side [tex]AB[/tex] of [tex]\bigtriangleup{ABC}[/tex] intersects side [tex]BC[/tex] at point [tex]D[/tex] and the perimeter of  [tex]\bigtriangleup{ACD}[/tex].

From the figure,

[tex]AE=BE[/tex]         .......(1)              (as [tex]DE[/tex] is perpendicular bisector of side [tex]AB[/tex])

Now, In [tex]\bigtriangleup{BED}[/tex] and [tex]\bigtriangleup{AED}[/tex]

     [tex]AE=BE[/tex]                                     ( from equation 1 )

[tex]\angle {BED} =\angle {AED}[/tex]                               ( Both [tex]90^\circ[/tex] )

    [tex]ED=ED[/tex]                                     ( Common side)

[tex]\bigtriangleup{BED}\cong\bigtriangleup{AED}[/tex]                              ( by SAS congruence rule)

      [tex]BD=AD[/tex]    .........(2)                   (by CPCT)

As per question,

The perimeter of ∆ABC is with 12 cm larger than the perimeter of ∆ACD.

[tex]AB+BC+AC=AC+CD+AD+12[/tex]

         [tex]AB+BC=AD+CD+12\\AD+CD=BD+CD\\AB+BC=BC+12\\[/tex]

                   [tex]AB=12\rm\;{cm}[/tex]

Hence, the length of [tex]AB[/tex] is [tex]12\rm\;{cm}[/tex].

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Answer:

Two column proofs are organized into statement and reason columns. Each statement must be justified in the reason column. The reason column will typically include "given", vocabulary definitions, and theorems.

Therefore, what can be used as a reason in a two-column proof are:

Postulates

Definitions

Answer:

5836200.

Step-by-step explanation:

Given  :  5836197 .

To find : Round 5836197 to the nearest hundred.

Solution : We have given 5836197

Step 1 : First, we look for the rounding place which is the hundreds place.

Step 2 : Rounding place is 97.

Step 3 : 97  is greater than  50 then it would be rounded up mean next number to 97 would be increase to 1 and 97 become 00.

Step 4 : 5836200.

Therefore, 5836200.

To find a, b, c for the solution:

Let's start by writing down the expression for the function x(t) and its derivative:

We have:

x(t) = at² + bt + c
and
x'(t) = 2at + b

Using x' and x into the differential equation x′ + 2x = t² + 4t + 7 gives us:

2at + b + 2*(at² + bt + c) = t² + 4t + 7
Expanding this gives:
2at² + 2bt + b + 4at + 2c = t² + 4t + 7

By equating the coefficients of equivalent powers of t on both sides, we get three equations:

For t² :
2a = 1
So, a = 1/2

For t:
2b + 4a = 4
Substitute a = 1/2 into the equation gives b = 1 - 2 = -1

For the constant term:
b + 2c = 7
Substituting b = -1 gives c = 4.

So the solution is a = 1/2, b = -1, c = 4.

So the specific solution of this differential equation is given by x(t) = (1/2)t² - t + 4.

Answer:

B

Step-by-step explanation:

Answer:= 5/6

Step-by-step explanation:hope this helps

Answer: n=296

Step-by-step explanation: 9 x 23 + 3 x 39 - 28 + n = 296  

Hope this helps! <3

Answer:

19 members voted.

Step-by-step explanation:

Percentage problems can be solved by a rule of three.

25% of the 76 members of the club voted in the election. How many members voted?

So 76 is 100% = 1. How much is 0.25?

76 - 1

x - 0.25

[tex]x = 76*0.25[/tex]

[tex]x = 19[/tex]

19 members voted.

After factoring x from both terms, you can factor the difference of two squares.

= x(x² -64)

= x(x -8)(x +8)

_____

It is worth remembering the "speciall form" that is the difference of two squares:

... a² - b² = (a -b)(a +b)

Answer:

x(x -8)(x +8)

Step-by-step explanation:

Compute for the total balance:

total balance = $415 + $215 = $630

Then we compute for the total credit limit:

total credit limit = $1,000 + $750 = $1,750

The credit utilization would simply be the percentage ratio of total balance over total credit limit. That is:

credit utilization = ($630 / $1,750) * 100%

credit utilization = 36%

The zeros of the function y = -3(x - 2)² + 6 are x = 2 + √2 and x = 2 - √2, which represent the hot dog prices at which the hot dog stand makes $0.00 profit.

The zeros of a function are the values of x that make the y-coordinate equal to zero. In this case, the function y = -3(x - 2)² + 6 represents the daily profit, and we need to find the x-values that result in a profit of $0.00. Setting the profit, y, to zero and solving for x, we get:

0 = -3(x - 2)² + 6

Adding 3(x - 2)² to both sides and simplifying the equation gives:

3(x - 2)² = 6

Dividing both sides by 3, we have:

(x - 2)² = 2

Take the square root of both sides, remembering to consider both the positive and negative square roots:

x - 2 = ±√2

Adding 2 to both sides gives us the final solutions:

x = 2 ± √2

So, the zeros of this function are x = 2 + √2 and x = 2 - √2. These are the hot dog prices at which the hot dog stand makes $0.00 profit.

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Answer: There are 95 students who scored between 35 and 61.

Step-by-step explanation:

Since we have given that

The following data : 11,35,61,70,79.

So, the median of this data would be = 61

First two data belongs to "First Quartile " i.e. Q₁

and the second quartile is the median i.e. 61.

The last two quartile belongs to "Third Quartile" i.e. Q₃

And we know that each quartile is the 25th percentile.

And we need "Number of students who scored between 35 and 61."

So, between 35 and 61 is 25% of total number of students.

So, Number of students who scored between 35 and 61 is given by

[tex]\dfrac{25}{100}\times 380\\\\=\dfrac{1}{4}\times 380\\\\=95[/tex]

Hence, There are 95 students who scored between 35 and 61.

The number of students who scored between 35 and 61 is 95

The 5 number summary is the value of the ;

The total Number of students = 380

The lower quartile (Lower 25%) = 35

The median (50%) = 61

This means that 25% of the total students scored between 35 and 61.

Hence, 95 students scored between 35 and 61.

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The derivative[tex]\( \frac{dw}{dt} \) is \( \frac{7t^6 e^{4-t}}{2+9t} - \frac{t^7 e^{4-t}}{2+9t} - \frac{9t^7 e^{4-t}}{(2+9t)^2} \).[/tex]

To find [tex]\( \frac{dw}{dt} \)[/tex] using the chain rule for the given function[tex]\( w = \frac{x e^y}{z} \), where \( x = t^7 \), \( y = 4 - t \), and \( z = 2 + 9t \)[/tex], follow these steps:

1. **Express ( w ) in terms of ( t ):**

  Substitute ( x ), ( y ), and ( z ) into ( w ):

[tex]\[ w = \frac{x e^y}{z} = \frac{(t^7) e^{(4 - t)}}{2 + 9t} \][/tex]

2. **Apply the chain rule:**

  The chain rule states that for a function ( w(t) ) defined implicitly by ( w = f(x(t), y(t), z(t)) ), the derivative [tex]\( \frac{dw}{dt} \)[/tex] is given by:

[tex]\[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt} \][/tex]

3. **Compute partial derivatives of ( w ) with respect to ( x ), ( y ), and ( z ):**

[tex]\( \frac{\partial w}{\partial x} = \frac{e^y}{z} \)[/tex]  

  [tex]\( \frac{\partial w}{\partial y} = \frac{x e^y}{z} \)[/tex]  

[tex]\( \frac{\partial w}{\partial z} = -\frac{x e^y}{z^2} \)[/tex]

4. **Compute [tex]\( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \):**[/tex]

[tex]\( \frac{dx}{dt} = 7t^6 \)[/tex]  

[tex]\( \frac{dy}{dt} = -1 \)[/tex]

[tex]\( \frac{dz}{dt} = 9 \)[/tex]

5. **Substitute these into the chain rule formula:**

[tex]\[ \frac{dw}{dt} = \frac{e^y}{z} \cdot 7t^6 + \frac{x e^y}{z} \cdot (-1) + \left(-\frac{x e^y}{z^2}\right) \cdot 9 \][/tex]

6. **Substitute[tex]\( x = t^7 \), \( y = 4 - t \), \( z = 2 + 9t \)[/tex] into the expression:**

[tex]\( e^y = e^{4 - t} \)[/tex]

  Substitute these values into the formula for [tex]\( \frac{dw}{dt} \):[/tex]

[tex]\[ \frac{dw}{dt} = \frac{e^{4 - t}}{2 + 9t} \cdot 7t^6 - \frac{t^7 \cdot e^{4 - t}}{2 + 9t} - \frac{9t^7 \cdot e^{4 - t}}{(2 + 9t)^2} \][/tex]

Therefore, [tex]\( \frac{dw}{dt} \)[/tex] is:

[tex]{\frac{dw}{dt} = \frac{7t^6 e^{4 - t}}{2 + 9t} - \frac{t^7 e^{4 - t}}{2 + 9t} - \frac{9t^7 e^{4 - t}}{(2 + 9t)^2} } \][/tex]

The coordinate of point C on line CE will be (8, 4).

What is Coordinates?

A pair of numbers which describe the exact position of a point on a cartesian plane by using the horizontal and vertical lines is called the coordinates.

Given that;

D is the midpoint of CE.

E has coordinates (-3,-2), and D has coordinates (2 1/2, 1).

Now, By the definition of midpoint;

Let the coordinate of point C = (x, y)

Then,

((x + (-3))/2 , (y + (-2))/2) = (2 1/2, 1)

By comparison we get;

x + (-3) / 2= 2 1/2

x - 3 = 2 (5/2)

x - 3 = 5

x = 3 + 5

x = 8

And, (y + (-2))/2 = 1

y - 2 = 2

y = 2 + 2

y = 4

Thus, The coordinate of point C = (x, y) = (8, 4)

So, The coordinate of point C on line CE will be (8, 4).

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The reduction factor of 0.8 means the lengths in the reduced triangle are 0.8 times those of the original.

Then the original 6 inch length is reduced to 0.8×6 inches = 4.8 inches in the reduced triangle.

Answer:

4.8 inches

Step-by-step explanation:

The scale factors are used to convert a figure into another one with similar characteristics but different lengths, in this example is a triangle, and in order to calculate the measure of the height you just have to multiply the original height by the scale factor:

6 inches * scale factor

6 inches* 0.8= 4.8 inches

So the resultant triangle will have a height of 4.8 inches.

sin(a+b) = -1/7 +√455/42 = 0.8721804464845457

cos(a-b) = √13/42 - √35/7 =  -0.7761476987942811

tan(a+b )= (6√13/13 + √35) / (1 - 6√455/13) = -0.525

Given sin(a) = 6/7 and cos(b) = -1/6, with a in quadrant II and b in quadrant III, we need to utilize trigonometric identities to find sin(a+b), cos(a-b), and tan(a+b).

Firstly, since a is in quadrant II, cos(a) is negative. We use the identity sin²(a) + cos²(a)=1 to find cos(a):

Similarly, since b is in quadrant III, sin(b) is also negative. We use the identity sin²(b) + cos²(b)=1 to find sin(b):

Now we can use the angle addition and subtraction formulas:

1. sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

2. cos(a - b) = cos(a)cos(b) + sin(a)sin(b)

3. tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))