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4 years ago

Given four functions , which one will have the smallest y-interception

Mathematics

1 answer:

Sedaia [141]4 years ago

4 0

The four y-intercepts are {$6, $2, $4, $10}. Which of these is the smallest?

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Ana uses Talk-A-Lot and Peter uses Talk-to-Me. Each company charges

mr Goodwill [35]

Answer:

They both talk for a different amount of time

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3 years ago

PLEASE HELP!!!! Nicholas is buying shirts for his baseball team. He will pay $8.75 for each shirt plus a one-time fee of $25 for

Marizza181 [45]

Answer:

y = 8.75x + 25

Step-by-step explanation:

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3 years ago

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∠A=5x−18. ∠B=3x+42∘ solve for x and then find the measure of ∠A

KatRina [158]

Answer:

x = 30 ∠A = 132 degrees

Step-by-step explanation:

Set the equations equal to each other to solve for x

5x - 18 = 3x + 42

-3x - 3x Subtract 3x from both sides

2x - 18 = 42

+ 18 + 18 Add 18 to both sides

2x = 60 Divide both sides by 2

x = 30

Now plug this into the equation for angle A

∠A = 5(30) − 18 Multiply

∠A = 150 - 18 Subtract

∠A = 132 degrees

If this answer is correct, please make me Brainliest!

6 0

3 years ago

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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

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4 years ago

Evaluate the expression 3(7 + 4)2 − 14 ÷ 7. (5 points)

Kitty [74]

<span>3(7 + 4)2 − 14 ÷ 7

First do the parenthesis

7 + 4 = 11

so the question looks like: 3(11)2 - 14/7

Then multiply 3, 11, and 2 together

3 x 11 x 2 = 66

66 - 14/7

14/7 equals to 2

66 - 2

Then just simplify

66 - 2 = 64

64 is your answer

hope this helps</span>

3 0

3 years ago

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