Find the area of the triangle with vertices: q(3,-4,-5), r(4,-1,-4), s(3,-5,-6).

The perimeter of the rectangle is 60 yards

What is the Perimeter of a Rectangle?

The perimeter P of a rectangle is given by the formula, P=2 ( L + W) , where L is the length and W is the width of the rectangle.

Perimeter P = 2 ( Length + Width )

Given data ,

Let the length of the rectangle be = a

Let the width of the rectangle be = b

The length of the rectangle is twice the width

So ,

a = 2b

The area of the rectangle is A = 200 yards²

The area of the rectangle is given by

Area of Rectangle = Length x Width

And , substituting the values for length and width , we get

Area of Rectangle = 2b x b

2b² = 200

Divide by 2 on both sides , we get

b² = 100

Taking square root on both sides , we get

b = 10

a = 2b

a = 20

Therefore , the length of the rectangle is 20 yards and width of the rectangle is 10 yards

Now , the perimeter of the rectangle is P = 2 ( length + width )

Perimeter of the rectangle P = 2 ( 10 + 20 )

                                                = 2 x 30

                                                = 60 yards

Hence , the perimeter of the rectangle is 60 yards

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Answer:

there are 13 more cups

Step-by-step explanation:

i already did it

Answer:

c: 18

Step-by-step explanation:

9.95 + 0.30s=c

 replaces s with 35

9.95 +0.30(35) =c

9.95 + 10.50 = 20.45

 it will cost $20.45

Answer:

[tex]P(n)=66n-1200[/tex]

Step-by-step explanation:

Let x be the revenue collected from selling each ticket.

We have been given that a charity organisation had to sell a few tickets just to cover necessary production costs of $1200. After selling 200 tickets, they had a net profit of $12,000.

Since we know that net profit is the difference between total revenue and total cost.

[tex]\text{Net profit}=\text{Total revenue- Cost}[/tex]

We can represent our given information in an equation to find the revenue collected from each ticket:

[tex]12000=200x-1200[/tex]

Let us solve for x by adding 1200 to both sides of equation.

[tex]12000+1200=200x[/tex]

[tex]13200=200x[/tex]

[tex]x=\frac{13200}{200}[/tex]

[tex]x=66[/tex]

Therefore, the revenue collected from selling each ticket is $66.

Now let us write the net profit, P(n), from fundraiser as the function of number of tickets sold,n.

The revenue from selling n tickets will be 66n.

Production costs = 1200

[tex]P(n)=66n-1200[/tex]

Therefore, our desired function will be: [tex]P(n)=66n-1200[/tex].

We have that the standard form of 24 ten thousands is

240,000

From the Question we are told that

24 ten thousands is given

Where

24 ten thousands=24*10000=240,000

Generally the standard form of 24 ten thousands is

240,000

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Jane studied a total of 5 11/12 hours for her history final and math exam.

To find the total number of hours Jane studied for the two tests, we add together the number of hours she studied for each test. Jane studied 4 2/3 hours for her history final and 4 1/4 hours for her math exam.

To add these fractions, we need to have a common denominator. The least common multiple of 3 and 4 is 12. We can convert 2/3 to 8/12 and 1/4 to 3/12.

Now we can add the fractions: 4 8/12 + 4 3/12 = 5 11/12.

Therefore, Jane studied a total of 5 11/12 hours for the two tests.

Final answer:

Jane studied a total of 7 3/4 hours for her history and math exams.

Explanation:

To find out how many hours Jane studied for the two tests, we simply need to add the hours she studied for each test together.

Jane studied 4 2/3 hours for her history final and 4 1/4 hours for her math exam.

We can add these two fractions by finding a common denominator and then adding the numerators.

In this case, the common denominator is 12. So, we have (4 * 4 + 2) / 3 = 18/3 and (4 * 3 + 1) / 4 = 13/4.

Adding these fractions together, we get (18/3) + (13/4) = 54/12 + 39/12 = 93/12. Simplifying this fraction, we get 7 3/4 hours.

Therefore, Jane studied a total of  7 3/4 hours for the two tests.

look where the line is on the y axis at x =2

 answer is y=-3

value is -3

  Both the triangles ΔABC and ΔADC are congruent by SSS postulate of congruence.

From the picture attached,

In ΔABC and ΔADC,

AB ≅ AD (Given)

BC ≅ CD (Given)

AC ≅ AC (Reflexive property)

Since, corresponding sides of ΔABC are equal to ΔADC,

Therefore, ΔABC ≅ ΔADC (By S-S-S postulate of congruence)

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the answer is F.The addition of zero postulate

The addition inverse postulate explains the relationship between numbers like 5 and -5 that add up to zero, crucial for understanding real number properties.

The addition inverse postulate is illustrated by the expression 5 + (-5) = 0 for real numbers. This postulate states that for every number, there exists an additive inverse that when added together results in zero.

For example, in the given expression, 5 and -5 are additive inverses of each other, and when added, they cancel out to give zero, satisfying the addition inverse postulate.

This concept plays a crucial role in understanding the properties and operations of real numbers.

Answer:

Step-by-step explanation:

Non-foliated metamorphic rocks are those which don't have a rough texture, like foliated ones, their composition gives them this characteristics: not having layers or banded appearance.

To prove that there exists a unique real number solution to the equation x³ + x² - 1 = 0 between x = 2/3 and x = 1, we can use the Intermediate Value Theorem and the property of the derivative.

To prove that there exists a unique real number solution to the equation x³ + x² - 1 = 0 between x = 2/3 and x = 1, we can use the Intermediate Value Theorem.

First, we substitute x = 2/3 into the equation and get a negative value (-1/27). Then, we substitute x = 1 into the equation and get a positive value (1).

Since the function is continuous between x = 2/3 and x = 1, and it changes sign, there must exist a solution somewhere between them.

To show that the solution is unique, we can use the fact that the derivative of the function, 3x² + 2x, is always positive for all real numbers.

This implies that the function is strictly increasing, so it can only intersect the x-axis at one point.

The correct equation to predict the number of free weight users is option B) p = 43.2(1.06)t. This equation accounts for a 6% annual increase in users.

Predicting the Number of Free Weight Users

In 1997, there were 43.2 million people using free weights. Given that the annual increase rate is 6%, we use the exponential growth formula to predict the number of people in future years. The correct equation to use is:

p = 43.2(1.06)t

This equation represents an exponential increase, where 'p' is the number of people using free weights and 't' is the number of years since 1997. This can be explained by the fact that a 6% annual increase multiplies the original number of users by 1.06 each year.

The answer to this question is 1

Answer:

B

Step-by-step explanation:

Answer:

c

Step-by-step explanation:

Answer:

[tex] \displaystyle10 \tan^{-1}( \sqrt {x}^{ } ) + \rm C[/tex]

Step-by-step explanation:

we would like to integrate the following integration:

[tex] \displaystyle \int \frac{5}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{3}{2} } } dx[/tex]

in order to do so we can consider using u-substitution

let our

[tex] \displaystyle u = {x}^{ \frac{1}{2} } \quad \text{and} \quad du = \frac{ {x}^{ - \frac{1}{2} } }{2} [/tex]

to apply substitution we need a little bit arrangement

multiply both Integral and integrand by 2 and ½

[tex]\displaystyle 2\int \frac{1}{2} \cdot\frac{5}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{3}{2} } } dx[/tex]

factor out [tex]x^{\frac{1}{2}}[/tex]:

[tex]\displaystyle 2\int \frac{1}{2 {x}^{ \frac{1}{2} } } \cdot\frac{5}{ (1+ {x}^{ } )} dx[/tex]

recall law of exponent:

[tex]\displaystyle 2\int \frac{ {x}^{ - \frac{1}{2} } }{2 } \cdot\frac{5}{ (1+ {x}^{ } )} dx[/tex]

apply substitution:

[tex]\displaystyle 2\int \frac{5}{ 1+ {x}^{ } } du[/tex]

rewrite x as [tex]\big(x^{\frac{1}{2}}\big)^{2}[/tex]:

[tex]\displaystyle 2\int \frac{5}{ 1+ ( {x ^{ \frac{1}{2} } })^{ 2 } } du[/tex]

substitute:

[tex]\displaystyle 2\int \frac{5}{ 1+ ( u)^{ 2 } } du[/tex]

recall integration rule of inverse trig:

[tex] \displaystyle 2 \times 5 \tan^{-1}(u)[/tex]

simplify multiplication:

[tex] \displaystyle10 \tan^{-1}(u)[/tex]

substitute back:

[tex] \displaystyle10 \tan^{-1}( {x}^{ \frac{1}{2} } )[/tex]

simplify if needed:

[tex] \displaystyle10 \tan^{-1}( \sqrt{x} )[/tex]

finally we of course have to add constant of integration:

[tex] \displaystyle10 \tan^{-1}( \sqrt {x}^{ } ) + \rm C[/tex]

and we are done!