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eduard
2 years ago
6

Find x in the given figure if m∠AOC = 124°.

Mathematics
1 answer:
Nastasia [14]2 years ago
3 0

Answer:

A) 8

Step-by-step explanation:

Add angles AOB and BOC to get angle AOC.

Since we know the value of AOC is 124 and AOB+BOC=AO

Then (8x+25)+(6x-13)=124.

Combine like terms to get 14x+12= 124.

Solve for x by subtracting 12 from 124, then divide the answer by 14.

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Emily wants to hang a painting in a gallery. The painting and frame must have an area of 31 square feet. The painting is 5 feet
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Area=legnth times width=31
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2 years ago
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In the solution, 3b + 9c = 3(b + 3). What should be done to make it CORRECT?
stiks02 [169]

Answer:

The second term of the binomial factor should be 3c instead of 3.

Step-by-step explanation:

Distributive property:

a*b + a*c = a*(b +c)

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2 years ago
Refer to the figure and find the volume V generated by rotating the given region about the specified line. R3 about AB.
mylen [45]

Answer: V = \frac{34}{45} \pi  


Explanation:

In the given system of coordinates OXY, the region R₃ is bounded by two functions:

y₁ = 2\sqrt[4]{x}  (green line)

y₂ = 2x  (blu line)

in the intervals:

0 ≤ x ≤ 1

0 ≤ y ≤ 2


We need to find the volume of this region rotated about the line AB, which is x = 1. In order to do so, we need to change system of coordinates, such as the rotation is about the y-axis, therefore we need to perform a translation:

\left \{ {{X=x+1} \atop {Y=y}} \right.

After the translation R₃ will be bounded by:

y₁ = 2\sqrt[4]{x+1}

y₂ = 2x + 2

in the intervals:

-1 ≤ x ≤ 0

0 ≤ y ≤ 2


At this point, we can use the washer method (see picture attached). The general formula is:

A = π(R² - r²)

where:

A = area

R = outer radius of a washer

r = inner radius of a washer


Since the radii are x-values which vary with the height, represented by the y-values, we need to write the inverse functions:

R: x_{1} = \frac{1}{16} y^{4} - 1 \\ r: x_{2} = \frac{1}{2} y - 1

[Note: I used the curves on the left side of the graph, but you could find the ones representing the right side of the graph and use those]


Now, we can find the function for the area of each washer:

A(y) = \pi [(\frac{1}{16}y^{4} - 1)^{2} - (\frac{1}{2}y - 1)^{2} ] \\ = \pi [\frac{1}{256}y^{8} - \frac{1}{8} y^{4} - \frac{1}{4} y^{2} + y ]


Therefore the volume of the region R₃ will be:

V = \int\limits^{y_{2}}_{y_{1}} {A(y)} \, dy

= \int\limits^2_0 {\pi [\frac{1}{256}y^{8} - \frac{1}{8}y^{4} - \frac{1}{4} y^{2} + y] } \, dy

= \pi [ \frac{1}{2304}y^{9} - \frac{1}{40}y^{5} - \frac{1}{12} y^{3} + \frac{1}{2} y^{2}]^{2}_{0}

= \frac{34}{45} \pi

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