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3 years ago

PLEASE HELP! I NEED ASAP!!!! Explain how to construct a perpendicular bisector

Mathematics

1 answer:

Anna007 [38]3 years ago

6 0

So the first step is to have a line segment.
Put your compass on one of the points and extend the compass to be slightly more than half. Just a bit over half.

Mark a small curvy line above your segment and on the bottom at that length.

Now pick up the compass, don't change the length of extension, and put it on the other point of your segment.

Mark above and below your segment like before so it can intersect.

draw a line through the two intersections.

I hope I answered your question.

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Find the common ratio for the geometric sequence for which a1=3 and a5=48

Rama09 [41]

The ratio is common, so you could write:

a1 * r^4 = a^5 (you do r^4 because there are 4 times you need to multiply by the ratio to get from a1 to a5)

Plug in the values:

3 * r^4 = 48

Divide by 3:

r^4 = 16

Take the fourth root of both sides:

r = 2

The common ratio is 2.

7 0

3 years ago

Read 2 more answers

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

Which of the following is the point-slope form of the line?

ryzh [129]

Answer:

Step-by-step explanation:

We can see that the slope is positive, which means that the x-term must be positive.

If you expand and simply both equations into the form y = mx + b, you will find that m is positive for A and negative for B, hence A is the correct answer.

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2 years ago

What is the point of intersection of the lines y=3/4x-1 and y=1/8x+4?

lesya [120]

Answer:

the answer for this question is option b

6 0

2 years ago

Original price: $35<br> Percent of discount:<br> Sale price: $31.50

AlladinOne [14]

Answer:

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Step-by-step explanation:

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3 years ago