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Papessa [141]
3 years ago
8

In ΔKLM, the measure of ∠M=90°, the measure of ∠K=78°, and KL = 35 feet. Find the length of MK to the nearest tenth of a foot.

Mathematics
2 answers:
Illusion [34]3 years ago
5 0

Answer:7.3

Step-by-step explanation:

Margaret [11]3 years ago
4 0

Answer:

7.3 feet

Step-by-step explanation:

osK=

hypotenuse

adjacent

​

=

35

x

​

\cos 78=\frac{x}{35}

cos78=

35

x

​

35\cos 78=x

35cos78=x

Cross multiply.

x=7.2769\approx \mathbf{7.3}\text{ feet}

x=7.2769≈7.3 feet

Type into calculator and round to the nearest tenth of a foot.

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Step-by-step explanation:

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\bold{METHOD\ 2:}\\\\k(x)=x-2,\ h(x)=x^2+1\\\\(h+k)(x)=h(x)+k(x)=(x^2+1)+(x-2)=x^2+x-1\\\\(h+k)(2)-\text{put}\ x=2\\\\(h+k)(2)=2^2+2-1=4+2-1=6-1=5

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Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
nirvana33 [79]

Answer:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}

We need to convert the rate given into m^3/min and we got:

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

5 0
3 years ago
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