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3 years ago

10

Can you please solve 2 and 3

1 answer:

Murrr4er [49]3 years ago

7 0

I’m pretty sure that I’m pretty sure that question 2 is 1 minute and 40 seconds. And for number 3 I got B. 0.25 because if you look at where the line makes a direct cross between pounds and dollars that first one is 8 pounds for 2 dollars. If you divide 8 by 2 you will get 0.25. Hope this helped :)

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Solve the system of equations.<br> -10x+3y=5 <br> x=y-4

blsea [12.9K]

Answer:

x=y-4 -10x+3y=5

y=x+4 y=10/3x+5/3

x+4 =10/3x+5/3

x=1

y=1+4 y=10/3(1)+5/3

y=5 y=5

so

x=1

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(1,5)

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3 years ago

What is the inverse of the function f(x) = x + 3

xenn [34]

Answer:

the answer would be f(x)=x-3

Step-by-step explanation:

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3 years ago

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nadezda [96]

1: 8 faces and 9 with the base 9 vertices and 16 edges

2: 3 faces and 5 with the bases 6 vertices and 9 edges

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3 years ago

V varies inversely with T and V=12 when T=6. Which equation shows this relationship

hodyreva [135]

Answer:

D

Step-by-step explanation:

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3 years ago

What is the exact area and arc length of these sectors? Please helppp

Karolina [17]

<u>Answers with step-by-step explanation:</u>

1. Area of sector 1 = $\frac{90}{360} \times \pi \times 12^2 = 36\pi$

2. Area of sector 2 = $\frac{45}{360} \times \pi \times 19^2 = \frac{2527}{8} \pi$

3. Area of sector 3 = $\frac{270}{360} \times \pi \times 15^2 = \frac{675}{4} \pi$

4. Area of sector 4 = $\frac{270}{360} \times \pi \times 6^2 = 27 \pi$

5. Arc length of sector 1 = $\frac{90}{360} \times 2 \times \pi \times 12 = 6\pi$

6. Arc length of sector 2 = $\frac{315}{360} \times 2 \times \pi \times 19 = \frac{133}{4} \pi$

7. Arc length of sector 3 = $\frac{270}{360} \times 2 \times \pi \times 15 = \frac{45}{2}\pi$

8. Arc length of sector 4 = $\frac{270}{360} \times 2 \times \pi \times 6 = 9\pi$

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3 years ago