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andrew-mc [135]

3 years ago

11

Please, help me out!

1 answer:

nasty-shy [4]3 years ago

8 0

A = lw + πr²
A = (125)(56) + (3.14)(28)²
A = 7000 + 3.14(784)
A = 7000 + 2461.76
A = 9461.76 yd²

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An inlet pipe on a swimming pool can be used to fill the pool in 40 hours. The drain pipe can be used to empty the pool in 42 ho

statuscvo [17]

Answer:

Pool will be filled in 280 hours

Step-by-step explanation:

Inlet pipe fills in 40 hours = 1 pool

Inlet pipe fills in 1 hours = $\frac{1}{40}$

Drain pipe empty in 42 hours = 1 pool

Drain pipe empty in 1 hour = $\frac{1}{42}$

If both pipes are opened together

then in pool fills in 1 hour = $\frac{1}{40}$ - $\frac{1}{42}$

on simplifying the right side ,we get $\frac{42-40}{(40)(42)}$

= $\frac{2}{(40)(42)}$

= $\frac{1}{840}$

$\frac{1}{840}$ pool fills in 1 hour

1 pool will be filled in 840 hours

$\frac{2}{3}$ pool is filled

empty pool = 1 - $\frac{2}{3}$ = $\frac{1}{3}$

therfore $\frac{1}{3}$ pool will be filled in $\frac{1}{3}$ X 840 =

= 280 hours

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3 years ago

PLEASEE HELP I ONLY HAVE 3 MINUTES LEFT PLEASE HELP ME

skelet666 [1.2K]

Answer:The graph of F is shifted 3 to the left and 1 up

Step-by-step explanation:

On the x axis, negative is left, positive is right. On the y axis, negative goes down, positive goes up. X is negative, it goes to the left, and y is positive, it moves up. Hope this helps and good luck!

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3 years ago

Yx(y+1) + 4x(y + 1) - 5(y + 1) factor

zheka24 [161]

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There are three containers filled with different gases as shown.

Natali [406]

$\qquad\qquad\huge\underline{{\sf Answer}}$

As we know,

$\sf \qquad density = \dfrac{mass}{volume}$

So, we can infer that :

$\sf \qquad mass= {density} \cdot{volume}$

Now, let's calculate the mass of gases in each case :

<h3>Case A : Hydrogen ~</h3>

$\qquad \sf \dashrightarrow \:mass = 0.09 \times (15) {}^{3}$

$\qquad \sf \dashrightarrow \:mass = 0.09 \times 3375 {}^{}$

$\qquad \sf \dashrightarrow \:mass = 303.75 \: \: mg {}^{}$

or

$\qquad \sf \dashrightarrow \:mass =0. 30375 \: \: g {}^{}$

<h3>Case B : Helium ~ </h3>

$\qquad \sf \dashrightarrow \:mass =0. 175 \: \cdot \: (14 \sdot 12 \sdot10)$

$\qquad \sf \dashrightarrow \:mass =0. 175 \: \cdot \: 1680$

$\qquad \sf \dashrightarrow \:mass =0. 175 \: \cdot \: 294 \: \: mg$

or

$\qquad \sf \dashrightarrow \:mass = 0.294 \: \: g$

<h3>Case C : Nitrogen ~ </h3>

$\qquad \sf \dashrightarrow \:mass = 1.251\: \cdot \: \dfrac{4}{3} \cdot3.14 \cdot(4) {}^{3}$

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or

$\qquad \sf \dashrightarrow \:mass = 0.335 \: \: g$

So, the arrangement of masses from least to greatest is :

(1.) Hydrogen < (2.) Helium < (3.) Nitrogen

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3 years ago

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scoray [572]

I think it is i dont know

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4 years ago