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VladimirAG [237]
3 years ago
8

What is the area of the irregular figure below?

Mathematics
2 answers:
expeople1 [14]3 years ago
7 0

Answer:

360 centimeters squared

Step-by-step explanation:

Liula [17]3 years ago
4 0

Answer:

360 centimeters squared.

Step-by-step explanation:

I did the test

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Two circular cylinders are similar. The ratio of the areas of their bases is 9:4. Find the ratio of the volumes of the similar s
OleMash [197]

Answer:

27 : 8

Step-by-step explanation:

Given 2 similar figures with ratio of sides a : b, then

ratio of areas = a² : b² and

ratio of volumes = a³ : b³

Here the ratio of areas = 9 : 4, thus

ratio of sides = \sqrt{9} : \sqrt{4} = 3 : 2

ratio of volumes = 3³ : 2³ = 27 : 8

8 0
3 years ago
Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48feet above the ground,
My name is Ann [436]

Answer: t = 3seconds

Step-by-step explanation:

To find the zero means we will equate the function to zero and then solve, that is

-16t^{2}+ 32t + 48 = 0

by factorizing , we have

(t+1)(-16t+48) = 0

Therefore:

t = -1 or t = 3

since time can not be negative , then , t = 3 seconds

5 0
3 years ago
Read 2 more answers
Ted needs an average of at least 70 on his three history tests. He has already scored 85 and 60 on two tests. What is the minimu
taurus [48]
To find the mean you must add up all the numbers and divide by how many there are:
35 + 60 + x
_________  =    70
        3

35 + 60 + x = 210

x = 65

He must score at least a 65 for a 70 average.
6 0
4 years ago
Read 2 more answers
SHOW YOUR WORK, <br><br>PLEASE HELP!!!!!!!!!
Shkiper50 [21]
To find the inverse, we swap the variables y and x, then solve for the new y.

3a. y=\frac{3}{x-1}

Swapping the variables: x=\frac{3}{y-1}
Solving for y: x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}
The domain of this inverse is x ≠ 0.
3b. y=x^2-1

Swapping: x = y^2 - 1
Solving for y: y^2 = x + 1 \\ y = \sqrt{x+1}
The domain of this inverse is x ≥ -1.
3c. y=\sqrt[3]{\frac{x-7}{3}}
Swapping: x=\sqrt[3]{\frac{y-7}{3}}
Solving for y: x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7
The domain of this inverse is all real numbers.
4a. y=\frac{3}{x-1}, y=1+\frac{3}{x}
y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x
y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x

4c. y=\sqrt[3]{\frac{x-7}{3}}, y=3x^3+7
y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x
y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x



3 0
3 years ago
So 7x4=28 then 7 division 28 =4
Tatiana [17]
Answer
True

Explanation
Common sense
5 0
2 years ago
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