3 years ago

How do I find it out

1 answer:

4 0

The two equations are in the slope-intercept form, y = mx + b.

m is the slope and it is the rate of change.
b is the constant, and it is the initial value.

Eq. A: y = 3x + 4 ---> rate of change = m = 3; initial value = b = 4
Eq. B: y = 5x + 2 ---> rate of change = m = 5; initial value = b = 2

Now you can see which statement is true.

A. False
B. True
C. False
D. False

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Does anyone know how to solve this problem? I am stuck. It is using solving systems of equations

bearhunter [10]

They both would need to get two toppings to cost the same amount which would be $8.60

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4 years ago

If f(x) = 6x2 - 4 and g(x) = 2x + 2, find (f - g)(x).

Dafna11 [192]

Answer:

6x² - 2x - 6

Step-by-step explanation:

note (f - g)(x) = f(x) - g(x), thus

f(x) - g(x)

= 6x² - 4 - (2x + 2) ← distribute

= 6x² - 4 - 2x - 2 ← collect like terms

= 6x² - 2x - 6

7 0

3 years ago

Read 2 more answers

Victor went for 1 run lats week . it was 3.5 km long . terrell went for 3 runs last week . each run was 1,250m long . wich perso

valina [46]

1 km = 1,000m
Victor: 3.5 km = 3,500m
Terrel: 1,250m *3 = 3,750m

Terrel ran a greater total distance

7 0

3 years ago

Use the law of cosines to explain why c^2=a^2+b^2 for triangle ABC, where angle C is a right angle.

trapecia [35]

Let $c$ be the length of the hypotenuse in the right triangle $ABC$ , with $m\angle C=90^\circ$ for $\angle C$ , the angle opposite the hypotenuse.

By the law of cosines,

$c^2=a^2+b^2-2ab\cos C$

But $\cos90^\circ=0$ , so we end up with $c^2=a^2+b^2$ .

5 0

3 years ago

1. The sum of 2 integers is -1. Their product is -12. What are the integers?

trapecia [35]

Answer:

1. -4, 3 2. 6, -4

7 0

3 years ago