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3 years ago

A circular pool has a radius of 12 feet. To the nearest square foot, what is the area of the trampoline?

2 answers:

6 0

452.16 because pie is 3.14 and the formula for area is pie x radius squared and radius here is 12 so 3.14x12^2 is 452.16

svet-max [94.6K]3 years ago

4 0

Answer:

452 ft.²

Step-by-step explanation:

πr² = 3.14 x 12²

= 452.16

= 452 ft.² (to the nearest square foot)

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Radical 5* radical 2

Setler [38]

Answer:

$\sqrt{10}$

Step-by-step explanation:

Rule: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$

$\sqrt{5} \times \sqrt{2} = \sqrt{5 \times 2} = \sqrt{10}$

6 0

2 years ago

Simplify completely quantity x squared plus 4 x minus 45 all over x squared plus 10 x plus 9 and find the restrictions on the va

Brums [2.3K]

Answer:

A) Quantity x minus 5 over quantity x plus 1, where x≠-1 and x≠-9

Step-by-step explanation:

$\frac{x^2 + 4x - 45}{x^2 + 10x + 9}$

Simplifying the numerator first:

x² + 4x - 45 using the quadratic formula you get;

(x - 5)(x + 9)

Then simplifying the denominator x² + 10x + 9 using a quadratic formula you get;

(x + 1)(x + 9)

Dividing the numerator and denominator now gives;

$\frac{(x - 5)(x + 9)}{(x + 1)(x + 9)}$

Cancelling (x + 9) throughout leaves you with;

$\frac{x - 5}{x + 1}$

The only restrictions here is if x = 1 and 9 which will give an undefined answer.

7 0

4 years ago

Read 2 more answers

a racing car starts fron rest .if it gains an acceleration of 5m/second square in 10 second ,calculate the final velocity.

hjlf

Answer:

The final velocity would be 50 m/s.

Its by formula,

$a = \frac{v - u}{t}$

hope it helps...

3 0

3 years ago

A particle moves on a line away from its initial position so that after t hours it is s = 2t^2 +3t miles from its initial positi

Vaselesa [24]

Answer:

Average speed of the particle is 13 miles per hour.

Step-by-step explanation:

From the given question expression for distance traveled by the particle is

$s=2t^{2}+3t$

Then fro the time interval (1, 4) we have to calculate the average speed.

At t=1, s=2(1)²+3(1) =2+3 =5 miles

And at t = 4 s=2(4)²+3(4) = 2(16)+12 = 32+12 =44 miles

So displacement of the particle in the time span from 1 to hours.

s = (44-5) = 39 miles

And the times taken = (4-1) =3 hours

From the formula speed = displacement/(time taken for displacement)

V = 39/3 = 13 miles/hour

4 0

3 years ago

−2 2 21 +(−0.25÷(− 1 4 )−1.5÷(− 3 16 ))÷(−4 1 11 )

Setler79 [48]

Answer:

$-2\frac{2}{21}+(-0.25\div(-14)-1.5\div(-\frac{3}{16}))\div(-4\frac{1}{11})=-4.0551$

Step-by-step explanation:

Given : Expression $-2\frac{2}{21}+(-0.25\div(-14)-1.5\div(-\frac{3}{16}))\div(-4\frac{1}{11})$

To find : Simplify the expression ?

Solution :

Using BODMAS,

$-2\frac{2}{21}+(-0.25\div(-14)-1.5\div(-\frac{3}{16}))\div(-4\frac{1}{11})$

$=-\frac{44}{21}+(-0.25\div(-14)-1.5\div(-\frac{3}{16}))\div(-\frac{45}{11})$

$=-\frac{44}{21}+(\frac{0.25}{14}+\frac{1.5\times 16}{3})\div(-\frac{45}{11})$

$=-\frac{44}{21}+(\frac{0.25}{14}+\frac{24}{3})\div(-\frac{45}{11})$

$=-\frac{44}{21}+(\frac{0.25\times 3+24\times 14}{14\times 3})\div(-\frac{45}{11})$

$=-\frac{44}{21}+(\frac{336.75}{42})\div(-\frac{45}{11})$

$=-\frac{44}{21}-\frac{336.75\times 11}{42\times 45}$

$=-\frac{44}{21}-\frac{3704.25}{1890}$

$=\frac{-44\times 1890-3704.25\times 21}{21\times 1890}$

$=\frac{-160949.25}{39690}$

$=−4.0551$

Therefore, $-2\frac{2}{21}+(-0.25\div(-14)-1.5\div(-\frac{3}{16}))\div(-4\frac{1}{11})=-4.0551$

6 0

3 years ago