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3 years ago

Select all of the expressions that are each the expressions that are equivalent to 4(5*+2y)

Mathematics

1 answer:

meriva3 years ago

5 0

Answer:

a, c, e

Step-by-step explanation: those are the asnwers

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What percent of 40 is 120?

aliya0001 [1]

Answer:

the correct answer is 33.33%

5 0

3 years ago

Read 2 more answers

1. In the following diagram, UW bisects XV at Y. Which of the

aev [14]

Answer:

Statement 4 is correct

Step-by-step explanation:

Here, we want to select which statement is true based on the given diagram;

The statement that must be true is that Y is the midpoint of XV

This is because, by bisection , we mean dividing into 2 equal parts

The line UW has divided the line XV into two equal parts

So this mean that Y is the midpoint of the line XV

6 0

3 years ago

Find the average rate of change of the function below over the interval (-4,-1).

babymother [125]

Answer:

-2

Step-by-step explanation:

$\frac{f(b)-f(a)}{b-a}\\ \\=\frac{f(-1)-f(-4)}{-1-(-4)}\\ \\=\frac{-4-2}{-1+4}\\ \\=\frac{-6}{3}\\ \\=-2$

Therefore, the average rate of change over the interval [-4,-1] is -2.

5 0

2 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

Kim has fraction 1 over 2 cup of almonds. she uses fraction 1 over 8 cup of almonds to make a batch of pancakes. part a: how man

Vesna [10]

She can make 4 batches of pancakes

6 0

3 years ago