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2 years ago

Please help this is due today

Mathematics

2 answers:

Masja [62]2 years ago

6 0

Answer:

Step-by-step explanation:

OverLord2011 [107]2 years ago

4 0

Answer:

Explanation:

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(4,-2) and (24,-17) finding the slope from points

Romashka [77]

Use the equation for the slope:
$slope = \frac{y2-y1}{x2-x1}$
The two points are equal to (x1, y1) and (x2, y2).

So choose one of your points to be (x1, y1). Let's choose (4, -2). That means x1 = 4 and y1 = -2.
The other point would then be (x2, y2). So x2 = 24 and y2 = -17.

Now put these values in to the equation:
$Slope = \frac{-17-(-2)}{24-4} = \frac{-15}{20} = \frac{-3}{4}$

The slope is -3/4.

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3 years ago

The coordinate plane below represents a town. Points A through F are farms in the town.

kondaur [170]

For part a,
y < x
y > -2x
Point B: (3,1) plugin x=3, y=1 y < x 1 < 3
which is true, so point B satisfies the first inequalityYou can graph the inequality y > 2x+5 and see which schools are in the shaded region

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3 years ago

G(x)= 3/4x+6 What is the value of g(12)?

gizmo_the_mogwai [7]

Answer:

G(12) = 15

Step-by-step explanation:

When doing g(x) equations, all you do is substitute the value inside the parentheses to x in the equation.

So you get 3/4(12) + 6.

Then simplify to get g(12) = 15

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3 years ago

Which of the following expressions is equivalent to -9?

dlinn [17]

Answer:

A and C

Step-by-step explanation:

-3*3 = -9

-1*-9 = 9

-27/3 = -9

-9/-1 = 9

6 0

3 years ago

Read 2 more answers

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

2 years ago