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2 years ago

2

Mathematics

1 answer:

Zigmanuir [339]2 years ago

6 0

Answer:

The weight of the water in the pool is approximately 60,000 lb·f

Step-by-step explanation:

The details of the swimming pool are;

The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet

The depth of the pool = 5 feet

The density of the water in the pool = 60 pounds per cubic foot

From the question, we have;

The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g

The volume of water in the pool = Cross-sectional area × Depth

∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³

Acceleration due to gravity, g ≈ 32.09 ft./s²

∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N

266,196.089 N ≈ 60,000 lb·f

The weight of the water in the pool ≈ 60,000 lb·f

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3 years ago

Read 2 more answers

Solve the equation 2^(2x+1)+8=17×2^x.

myrzilka [38]

Oooh, looks fun

ok
erm
we might want to know some properties
$(x^m)(x^n)=x^{m+n}$
$(a^b)^c=a^{bc}$
if $a^m=a^n$ where a=a, then assume m=n

$2^{2x+1}=(2^{2x})(2^1)$
so
$(2^{2x})(2)+8=17(2^x)$
$(2^{2x})(2)+2^3=17(2^x)$
$(2^x)^2(2)+2^3=17(2^x)$
minus 17*2^x both sides
$(2^x)^2(2)+2^3-17(2^x)=0$
use u subsitution, u=2ˣ
$(u)^2(2)+2^3-17(u)=0$
solve $2u^2+8-17u=0$
or
$2u^2-17u+8=0$
ac method
2 times 8=16
what 2 number multiply to get -17 and add to get 16
-16 and -1
2u²-1u-16u+8=0
(2u²-1u)+(-16u+8)=0
u(2u-1)+(-8)(2u-1)=0
(u-8)(2u-1)+0
u-8=0
u=8

2u-1=0
2u=1
u=1/2

now
u=2ˣ

8=2ˣ
$2^3=2^x$
3=x

and
$\frac{1}{2}=2^x$
$2^{-1}=2^x$
-1=x

x=-1 and 3
neat problem

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2 years ago