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matrenka [14]
3 years ago
13

How do you find the circumference of a circle?

Mathematics
1 answer:
irina1246 [14]3 years ago
6 0
It’s C=2πr to find the circumference of a circle
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How many solutions does the system have?
klio [65]

it have manny solution it depand on a method u are using

4 0
2 years ago
If you drive 300 miles and do 70 per hour how much minutes does it take​
dangina [55]

Answer: well I'm sorry I can't help that much but 300÷70(per hour)=4.28571428571 (Yes I did the math in my head) but it would be an estimate of 4 or 5 minutes. If I helped that's the least I can do. Your welcome! Anytime!!

Step-by-step explanation:

5 0
3 years ago
A weir is used to measure water flow in a channel. For a rectangular broad crested weir, the flow Q in cubic feet per second is
NikAS [45]

Answer:Find the water height for a weir that is 3 feet long and has flow of 38.4 cubic feet per second. SOLUTION: Therefore, the water height for the weir is 4 feet.

6 0
3 years ago
Read 2 more answers
A builder makes drainpipes that drop 1 cm over a
LUCKY_DIMON [66]

Answer:

700.4 cm

Step-by-step explanation:

This involves two similar triangles.

Both triangles are right triangles.

One has legs measuring 1 cm and 30 cm. We can find the hypotenuse by using the Pythagorean theorem.

(1 cm)^2 + (30 cm)^2 = c^2

c^2 = 901 cm^2

c = sqrt(901) cm

The second triangle has one leg with length 700 cm. This leg corresponds to the 30-cm leg in the other triangle. Since the triangles are similar, we can use a proportion to find the hypotenuse of the second triangle.

(30 cm)/(700 cm) = [sqrt(901) cm]/x

3/70 = sqrt(901) cm/x

3x = 70 * sqrt(901) cm

x = 70 * sqrt(901) cm/3

x = 700.4 cm

Answer: 700.4 cm

8 0
3 years ago
Read 2 more answers
PLEEEEEEEAAAAAAASSSSSSSSSSEEEEEEEEEEE help me!!!!!!!!!!
defon

Answer:

\boxed{\sf \ \ \ \dfrac{18}{(x-7)(x+12)(x+30)} \ \ \ }

Step-by-step explanation:

hello,

first of all we will study the quadratic expressions

we can write that, (the different answers provide good clues :-) )

x^2+5x-84=(x-7)(x+12)

and

x^2+23x-210=(x-7)(x+30)

so first of all as we cannot divide by 0 we need to take x different from 7, -12 and -30 and then we can write

\dfrac{1}{x^2+5x-48}-\dfrac{1}{x^2+23x-210}=\dfrac{1}{(x-7)(x+12)}-\dfrac{1}{(x-7)(x+30)}\\\\=\dfrac{(x+30) -(x+12)}{(x-7)(x+12)(x+30)}=\dfrac{x+30-x-12}{(x-7)(x+12)(x+30)}\\\\=\dfrac{18}{(x-7)(x+12)(x+30)}

hope this helps

7 0
3 years ago
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