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3 years ago

Unit 4: Linear Equations

Mathematics

1 answer:

Aleks [24]3 years ago

8 0

Do what the other person’s answer is :)

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Please help!

Lubov Fominskaja [6]

Answer:

The answer is d (-21/2 , 2)

Step-by-step explanation:

That's where the lines meet up

8 0

3 years ago

Find the slope between the points (−3,−5) and (10,-5)

balandron [24]

Answer:

Step-by-step explanation:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$m = \dfrac{-5 - (-5)}{-3 - 10}$

$m = \dfrac{0}{-13}$

$m = 0$

3 0

3 years ago

Read 2 more answers

the function h(t)=-16t^2+110t+72 models the height h, in feet, of an object above ground t seconds after being launched straight

Anettt [7]

Answer:

The initial height of an object above ground before being launched straight up in the air

Step-by-step explanation:

we have

$h(t)=-16t^{2}+110t+72$

we know that

The number 72 represent the y-intercept of the function

The y-intercept is the value of y when the value of x is equal to zero

In this problem

The value of h when the value of t is zero

Therefore

The initial height of an object above ground before being launched straight up in the air

7 0

3 years ago

Write an equivalent expression by distributing the " sign outside the parentheses: 5n - (7.4p - 9.6

liberstina [14]

5n - 7.4p + 9.6 is the answe

4 0

3 years ago

A particle moves along the curve y = 5x^2 – 1 in such a way that the y value is decreasing at the rate of 2 units per second. At

Shtirlitz [24]

Answer:

The correct option is;

Increasing one fifth unit/sec

Step-by-step explanation:

The equation that gives the curve of the particle of the particle is y = 5·x² - 1

The rate of decrease of the y value dy/dt = 2 units per second

We have;

dy/dx = dy/dt × dt/dx

dy/dx = 10·x

dy/dt = 2 units/sec

dt/dx = (dy/dx)/(dy/dt)

dx/dt = dy/dt/(dy/dx) = 2 unit/sec/(10·x)

When x = 1

dx/dt = 2/(10·x) = 2 unit/sec/(10 × 1) = 1/5 unit/sec

dx/dt = 1/5 unit/sec

Therefore, x is increasing one fifth unit/sec.

8 0

3 years ago