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alexandr402 [8]

2 years ago

11

In the figure, j∥k and m∠1 = 63°.

2 answers:

Volgvan2 years ago

8 0

If your in k12 the answer your looking for is

<u><em>: 117°</em></u> hope this short answer helps verify.

ANTONII [103]2 years ago

3 0

Answer:

∠ 4 = 117°

Step-by-step explanation:

∠ 1 and ∠ 4 are alternate angles and supplementary, sum to 180°, then

∠ 4 = 180° - 63° = 117°

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What is the value of d/dx (6/x^4 + 1/x^2) at x = -1 ?

zheka24 [161]

Answer:

\frac{d}{dx}\left(\frac{1+x^4+x^6}{x^2+x+1}\right)=\frac{4x^7+5x^6+8x^5+3x^4+4x^3-2x-1}{\left(x^2+x+1\right)^2}

Step-by-step explanation:

7 0

2 years ago

Draw an example of a composite figure that has a volume between 750 cubic inches and 900 cubic inches

grigory [225]

Volume:

$V \approx 888.02in^3 \\ \\ And, \ 750in^3$

<h2>Explanation:</h2>

A composite figure is formed by two or more basic figures or shapes. In this problem, we have a composite figure formed by a cylinder and a hemisphere as shown in the figure below, so the volume of this shape as a whole is the sum of the volume of the cylinder and the hemisphere:

$V_{total}=V_{cylinder}+V_{hemisphere} \\ \\ \\ V_{total}=V \\ \\ V_{cylinder}=V_{c} \\ \\ V_{hemisphere}=V_{h}$

So:

$V_{c}=\pi r^2h \\ \\ r:radius \\ \\ h:height$

From the figure the radius of the hemisphere is the same radius of the cylinder and equals:

$r=\frac{8}{2}=4in$

And the height of the cylinder is:

$h=15in$

So:

$V_{c}=\pi r^2h \\ \\ V_{c}=\pi (4)^2(15) \\ \\ V_{c}=240\pi in^3$

The volume of a hemisphere is half the volume of a sphere, hence:

$V_{h}=\frac{1}{2}\left(\frac{4}{3} \pi r^3\right) \\ \\ V_{h}=\frac{1}{2}\left(\frac{4}{3} \pi (4)^3\right) \\ \\ V_{h}=\frac{128}{3}\pi in^3$

Finally, the volume of the composite figure is:

$V=240\pi+\frac{128}{3}\pi \\ \\ V=\frac{848}{3}\pi in^3 \\ \\ \\ V \approx 888.02in^3 \\ \\ And, \ 750in^3$

<h2>Learn more:</h2>

Volume of cone: brainly.com/question/4383003

#LearnWithBrainly

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2 years ago

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Bess [88]

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2 years ago

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BaLLatris [955]

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3 years ago

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gavmur [86]

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2 years ago

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