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andrew-mc [135]
3 years ago
11

Please, help me out!

Mathematics
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

6 0
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!

6 0
3 years ago
Yx(y+1) + 4x(y + 1) - 5(y + 1) factor
zheka24 [161]

( yx + 4x - 5 )( y + 1 )

.....................................

5 0
3 years ago
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}

\qquad \sf  \dashrightarrow \:mass = 0.417\:  \cdot  \: 803.84

\qquad \sf  \dashrightarrow \:mass = 335.201 \:  \: mg

or

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

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

  • (1.) Hydrogen < (2.) Helium < (3.) Nitrogen
5 0
3 years ago
The points p(a,b,c) and q(2,3,5) are symmetric about the point(4,0,1) find r
scoray [572]

 I think it is i dont know
3 0
4 years ago
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