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Troyanec [42]
3 years ago
8

No Docs/No Files Only real Answer 65 points and a 5-star rated I Will also follow back

Mathematics
2 answers:
dlinn [17]3 years ago
7 0

Well, If the area of ABC is 20cm², Then the area of ABS would be 10cm² and the area of ACS would also be 10cm². You basically are just cutting the ABC in half. I hope this helps!

DENIUS [597]3 years ago
5 0

Answer:

a) 400

b)160,000

Step-by-step explanation:

times 20 by 20 hence 20² for a. you get 400, for b, just do 400² which gets you 160,000.

If its wrong im so sorry

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kupik [55]

Answer:JI=16, JH=15, <W=37, <J=114, <I= 29

Step-by-step explanation:

When naming congruent triangles, the corresponding points are named in order from one triangle to another.

Since it states triangle WXY is congruent to triangle HJI this means that W corresponds to H, X corresponds to J and Y corresponds to I.

JI corresponds to XY, JH corresponds to XW angle W corresponds to angle H, angle J corresponds to angle X and angle I corresponds to angle Y. Use the triangle congruency statement as your roadmap.

To find the missing angle in the triangle, use the triangle angle sum theorem which states that all interior angles of a triangle sum to 180.

37+114+<I= 180. Combine like terms.

151+<I=180. Subtract 151 from both sides.

<I= 29

8 0
3 years ago
Every time a customer purchases a new computer, one month later they receive a survey in the mail. Eighty-five percent of the su
True [87]

Answer:

Respondents

Step-by-step explanation:

The customers that buy a new computer and receive a survey in the mail about the performance of the computer after one month are called respondents.

They are called respondents because they took part in a survey or poll about the performance of the computer they bought a month ago. Statistically, not everyone that got the mail took part in the poll, those ones are called non respondents, while the ones that took part in the survey are known as respondents.

4 0
3 years ago
Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle below.
AfilCa [17]

Answer:

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Step-by-step explanation:

The Pythagorean Theorem:

x^2+y^2=z^2

where x and y are two legs, and z is the hypotenuse

Plugging this in, we have

40^2 + 9^2 = 1681\\\sqrt{1681}=41

8 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%20%20%5Crm%5Csum%20%5Climits_%7Bn%20%3D%200%7D%5E%7B%20%5Cinfty%20%7D%20%20%5Carcsin%20%5Clar
sineoko [7]

Recall that over an appropriate domain,

\arcsin(x) \pm \arcsin(y) = \arcsin\left(x \sqrt{1-y^2} \pm y \sqrt{1-y^2}\right)

Let x=\frac1{n+1} and y=\frac1{n+2} (these belong to the "appropriate domain", so the identity holds). We have

\dfrac{\sqrt{n+3}}{(n+2)\sqrt{n+1}} = \dfrac1{n+1} \sqrt{\dfrac{(n+3)(n+1)}{(n+2)^2}} = \dfrac1{n+1} \sqrt{1 - \dfrac1{(n+2)^2}} = x \sqrt{1-y^2}

and

\dfrac{\sqrt n}{(n+1)\sqrt{n+2}} = \dfrac1{n+2} \sqrt{\dfrac{n(n+2)}{(n+1)^2}} = \dfrac1{n+2} \sqrt{1 - \dfrac1{(n+1)^2}} = y \sqrt{1-x^2}

Then the sum telescopes, as

\displaystyle \sum_{n=0}^\infty \arcsin\left(x \sqrt{1-y^2} - y \sqrt{1-x^2}\right) = \sum_{n=0}^\infty \left( \arcsin(x) - \arcsin(y) \right) \\\\ = \left(\arcsin(1) - \arcsin\left(\frac12\right)\right) + \left(\arcsin\left(\frac12\right) - \arcsin\left(\frac13\right)\right) + \cdots \\\\ = \arcsin(1) = \boxed{\frac\pi2}

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Schach [20]
The answer is 12 m.

Because you're not studying!

Next time do your mom.

3 0
3 years ago
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