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3 years ago

What is the area of the red zone? The red zone is shown in the picture

Mathematics

1 answer:

jarptica [38.1K]3 years ago

5 0

the top is equal to the bottom so u take the top and add it with it's self 24 +24 and do the same with the sides 9+9 add them together to get 66

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Square root of 512m^3. show work

ElenaW [278]

The solution would be like this for this specific problem:

√512m³ = √256m² × √2m

= 16 × √2m

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0

3 years ago

Translate the sentence into an inequality.

Luda [366]

Answer:

5 + c > -22

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Inequalities

Step-by-step explanation:

Step 1: Define

Sum of 5 and c is greater than -22

↓ Identify

Sum = addition

5 + c

Is greater than = inequality

Add them all together:

5 + c > -22

8 0

3 years ago

I WILL GIVE BRAINLIEST

Vaselesa [24]

Answer:

Step-by-step explanation:

80-225*1/3=80-75=5

3 0

3 years ago

Need help with this question -9×3-4^2/ 3^2 + 2^2

Vesnalui [34]

-9 × 3 - 4² = -27 - 16 = -43

3² + 2² = 9 + 4 = 13

(-9 × 3 - 4²)/(3² + 2²) = -43/13 = -3 4/13

3 0

3 years ago

A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate

Katyanochek1 [597]

Answer:

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

$\frac{dV}{dt}=4 \ m^3/s$

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is $V=\pi r^2 h$

$=( \pi \times 2^2\times h)\ m^3$

$\therefore V= 4\pi h$

Differentiating with respect to t.

$\frac{dV}{dt}=4\pi \frac{dh}{dt}$

Putting the value $\frac{dV}{dt}$

$\Rightarrow 4\pi \frac{dh}{dt}=4$

$\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}$

The rate change of height does not depend on the height.

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

4 0

3 years ago