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alexandr1967 [171]
2 years ago
7

Compare 2/5 and 8/10

Mathematics
1 answer:
ELEN [110]2 years ago
5 0

Answer:

2/5 is 4/10. 8/10 is 2 time greater than 2/5.

Step-by-step explanation:

4/10= 0.8, which can be written 8/10.

Make me brainlist.

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If the area of a circle is 58 square feet, find the circumference. A. 42.5 ft. B. 4.25 ft. C. 22.35 ft. D. 26.99 ft.
PSYCHO15rus [73]

Answer: OPTION D.

Step-by-step explanation:

The formula for calculate the circumference of a circle is:

C=2r\pi

Where <em>r</em> is the radius.

The formula for calculate the area of a circle is:

A=r^2\pi

Where <em>r</em> is the radius.

Solve for <em>r</em> from A=r^2\pi to calculate it:

r=\sqrt{\frac{A}{\pi}}\\r=\sqrt{\frac{58ft^2}{\pi}}\\r=4.296ft

Subsitute the radius into C=2r\pi. Then:

C=(2)(4.296ft)\pi=26.99ft

3 0
4 years ago
Read 2 more answers
Mary has three baking pans. Each pan is 8" × 8" × 3". Which expression will give her the total volume of the pans?
klio [65]
The volume of one pan is 192 units cubed. The volume of the three pans is 576 units cubed.
7 0
3 years ago
An exponential function f(x) = ab^xpasses through the points (0, 2) and (3, 54). What are the values of a and
Vsevolod [243]

Answers:

a = 2

b = 3

=======================================================

Explanation:

Plug in x = 0 and y = 2 to find that

y = a*b^x

2 = a*b^0

2 = a*1

2 = a

a = 2

Then plug in x = 3 and y = 54 to determine the value of b

y = a*b^x

y = 2*b^x

54 = 2*b^3

2b^3 = 54

b^3 = 54/2

b^3 = 27

b = (27)^(1/3)

b = 3

So we have y = a*b^x update to y = 2*3^x

7 0
3 years ago
What is the area of a square with side lengths of 3/5 units?
umka2103 [35]

Answer:

We must solve 3/5 x 3/5.

3/5 x 3/5 = 0.36

The units we'll use is  because units are the unt we're given, and it is squared because it is the area of a shape.

Your final answer is 0.36

6 0
3 years ago
Find the domain of the function.
timurjin [86]

When taking square roots, you can't take square roots of negative roots of negative numbers. So, what will work for the domain of u(x) is what makes u(x) zero or more. We can make an inequality for that.

u(x) ≥ 0.

\sqrt{9x+27} \geq  0

9x + 27 ≥ 0 by squaring both sides

9x ≥ -27

x ≥ -3

So the domain of the function is when x ≥ -3 is true.

5 0
4 years ago
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