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faltersainse [42]
3 years ago
7

Find X if m/_ UVH=50°, m/_HVW=25x-5, and m/_ UVW = 35x-5.​

Mathematics
1 answer:
ra1l [238]3 years ago
7 0

Step-by-step explanation:

first you have to solve for x

35x - 5=50

35-35x-5= 50-35

x-5=15

x- 5÷5=15÷5

x=3

Now we have to substitute x into the equation.

35×(3)-5=

175-5=170

angle UVW= 170

to find angle HUW instead of solving the equation just subtract 170 from 50 and you will get 120.

angle UVH=50

angle HUW=120

angle UVW=170

I hope this helps srry it took so long but gn and gl.

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If tan of angle theta = 11/60 , what is the value of cot angle theta?
UkoKoshka [18]
The tangent and cotangent functions of the angle are related such that one is the reciprocal of the other. From the given tan theta = 11 / 60. The reciprocal of this number is 60 / 11. This value is also the cotangent of the angle. Thus, the cotangent angle theta is 60 / 11. 
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3 years ago
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A typical running track is an oval with 74-m-diameter half circles at each end. a runner going once around the track covers a di
denpristay [2]

Answer:

Her centripetal acceleration during the turn at each end of the track is 0.432\, \frac{m}{s^{2}}

Step-by-step explanation:

Total distance covered in one round , D= 400 m

Time taken to cover one round , T = 1 min 40 s = 100 sec

Speed of runner , V=\frac{D}{T}=\frac{400}{100}\, \frac{m}{s}= 4.0\, \frac{m}{s}

Now centripetal acceleration is given by

a_c=\frac{V^{2}}{R}

where V= 4.0\frac{m}{s}

Radius, R= \frac{Diameter}{2}=\frac{74}{2}m=37\, m

\therefore a_c=\frac{4^{2}}{37}\, \frac{m}{s^{2}}=0.432\, \frac{m}{s^{2}}

Thus her centripetal acceleration during the turn at each end of the track is 0.432\, \frac{m}{s^{2}}

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4 years ago
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The mass of an object
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2 years ago
Hi! I need help with the attached question in calc. Thank you:)
Elena L [17]

Answer:

3π square units.

Step-by-step explanation:

We can use the disk method.

Since we are revolving around AB, we have a vertical axis of revolution.

So, our representative rectangle will be horizontal.

R₁ is bounded by y = 9x.

So, x = y/9.

Our radius since our axis is AB will be 1 - x or 1 - y/9.

And we are integrating from y = 0 to y = 9.

By the disk method (for a vertical axis of revolution):

\displaystyle V=\pi \int_a^b [R(y)]^2\, dy

So:

\displaystyle V=\pi\int_0^9\Big(1-\frac{y}{9}\Big)^2\, dy

Simplify:

\displaystyle V=\pi\int_0^9(1-\frac{2y}{9}+\frac{y^2}{81})\, dy

Integrate:

\displaystyle V=\pi\Big[y-\frac{1}{9}y^2+\frac{1}{243}y^3\Big|_0^9\Big]

Evaluate (I ignored the 0):

\displaystyle V=\pi[9-\frac{1}{9}(9)^2+\frac{1}{243}(9^3)]=3\pi

The volume of the solid is 3π square units.

Note:

You can do this without calculus. Notice that R₁ revolved around AB is simply a right cone with radius 1 and height 9. Then by the volume for a cone formula:

\displaystyle V=\frac{1}{3}\pi(1)^2(9)=3\pi

We acquire the exact same answer.

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3 years ago
a jeweler buys a ring for $85.00 and sell it with 135% increase in price. How much did he sell it for?
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Well if the increase is 135% then
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