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frutty [35]
2 years ago
8

Please help me with this math problem!! Will give brainliest!! :)

Mathematics
1 answer:
aleksandrvk [35]2 years ago
8 0

Answer:

-3

Step-by-step explanation:

(fg)(-1) is just f(-1) * g(-1). From the graph on the left, we can see that f(-1) is 1, since (-1,1) is on the graph. From the graph on the tight, we can see that g(-1) is -3, since (-1,-3) is on the graph. Now, we just have to multiply 1 and -3: 1*-3 = -3

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Convert the equation − x − 2 y = 8 from standard to slope-intercept form
Molodets [167]

Answer:

y = -1/2x - 4

Step-by-step explanation:

We have our equation in standard form: -x - 2y = 8. We can use simplifying rules to isolate y on one side of the equation and give us our y = mx + b format. I'll begin by adding x to both sides.

-x - 2y = 8

-2y = x + 8

Now that we have y on one side, we can divide everything by -2.

-2y/-2 = x/-2 + 8/-2

y = -1/2x - 4

That looks like y = mx + b form.

If you have any further questions or need clarification on anything, let me know!

4 0
3 years ago
Simplify.<br>y^5 ÷ y^3 × y^2 <br>NEED HELP ASAPP PLEASEEE ??? <br>Will get BRAINLIEST***​
Volgvan

Answer:

y^5  ÷  y^3 * y^2 = y^4

Step-by-step explanation:

https://www.symbolab.com/solver/factor-calculator/Simplify%20y%5E%7B5%7D%5Cdiv%20y%5E%7B3%7D%C3%97y%5E%7B2%7D%20

(use that link if you want to see the work for yourself)

:)

3 0
3 years ago
Linda and Colleen both leave the library at the same time, but in opposite direction. If Colleen travels 8mph faster than Linda
pshichka [43]
X - Linda's speed
(x+8 )- Colleen's speed

8x - Linda's distance for 8 h
8(x+8) = 8x+64  - Colleen's distance for 8 h

8x + 8x + 64 = 176 - distance between Colleen and Linda after 8 h


8x + 8x + 64 = 176 
16x +64 = 176
16x = 176 - 64
16x = 112
x= 7 mph Linda's speed
x+8
= 7+8 = 15 mph Colleen's speed
8 0
3 years ago
What is Pythagorean’s theorem??
Alchen [17]

Answer:

HOPE it's helpful

......

5 0
3 years ago
Read 2 more answers
Someone answer. Will give brainlest
Leto [7]

Area inside the semi-circle and outside the triangle is  (91.125π - 120) in²

Solution:

Base of the triangle = 10 in

Height of the triangle = 24 in

Area of the triangle = \frac{1}{2} bh

                                $=\frac{1}{2} \times 10\times24

Area of the triangle = 120 in²

Using Pythagoras theorem,

\text{Hypotenuse}^2=\text{base}^2+\text{height}^2

\text{Hypotenuse}^2=10^2+24^2

\text{Hypotenuse}^2=100+576

\text{Hypotenuse}^2=676

Taking square root on both sides, we get

Hypotenuse = 23 inch = diameter

Radius = 23 ÷ 2 = 11.5 in

Area of the semi-circle = \frac{1}{2}\pi r^2

                                      $=\frac{1}{2} \pi \times (13.5)^2

Area of the semi-circle = 91.125π in²

Area of the shaded portion = (91.125π - 120)  in²

Area inside the semi-circle and outside the triangle is  (91.125π - 120)  in².

5 0
3 years ago
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