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GarryVolchara [31]
3 years ago
13

PLEASE HELP WITH BOTH PROBLEMS. A.S.A.P DIRECTIONS ON PHOTO ^

Mathematics
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

6) 18^2+x^2=30^2\\324+x^2=900\\x=\sqrt{900-324} \\x=\sqrt{576} \\x=24



7) x^2+4^2=5^2\\x^2+16=25\\x=\sqrt{25-16}\\ x=\sqrt{9}\\x=3

Step-by-step explanation:


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Simplify the expression.Can you help me with this one. I'm stuck at 12^x4 - 15x^2
Blababa [14]

Simplify the given expression as shown below

\begin{gathered} 3x(4x^4-5x)=3x*4x^4+3x(-5x)=12x^5+(-15x^2) \\ =12x^5-15x^2 \end{gathered}<h2>Therefore, the answer is 12x^5-15x^2</h2>

3 0
1 year ago
A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lower base of 4 m, and an uppe
Ganezh [65]

Answer:

0.28cm/min

Step-by-step explanation:

Given the horizontal trough whose ends are isosceles trapezoid  

Volume of the Trough =Base Area X Height

=Area of the Trapezoid X Height of the Trough (H)

The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)

The Volume of water in the trough at any time

Volume=\frac{1}{2} (b_{1}+4+2x)h X H

Volume=\frac{1}{2} (4+4+2x)h X 16

=8h(8+2x)

V=64h+16hx

We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles

x/h=1/4

4x=h

x=h/4

Substituting x=h/4 into the Volume, V

V=64h+16h(\frac{h}{4})

V=64h+4h^2\\\frac{dV}{dt}= 64\frac{dh}{dt}+8h \frac{dh}{dt}

h=3m,

dV/dt=25cm/min=0.25 m/min

0.25= (64+8*3) \frac{dh}{dt}\\0.25=88\frac{dh}{dt}\\\frac{dh}{dt}=\frac{0.25}{88}

=0.002841m/min =0.28cm/min

The rate is the water being drawn from the trough is 0.28cm/min.

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The least common denominator is 70 because both 14 and 10 go into 70.

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Answer:

B

Step-by-step explanation:

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1 US dollar= 1.18 Canadian dollars
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