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lianna [129]
2 years ago
15

Heeeellpp!!!! ABC ≅

Mathematics
2 answers:
Andreyy892 years ago
8 0
Triangle ABC  is congruent to  triangle QPR.
Artemon [7]2 years ago
6 0
Triangle abc is congruent to triangle QPR
You might be interested in
I'm stuck on this homework question. I forgot the formula to find the side. This is 9th grade geometry.
dexar [7]

Answer:

x = 2root(22)

see image.

Step-by-step explanation:

The triangle shown is one big triangle cut into two more smaller triangles: one medium-sized and one smaller.

ALL THREE TRIANGLES ARE SIMILAR BY AA.

Set the two smaller triangles up so you can see the corresponding sides. x is the short leg in one triangle and it is the long leg in the smallest triangle. Set up a proportion.

22/x = x/4

crossmultiply

x^2 = 22•4

x^2 = 88

square root both sides.

x = sqroot(88)

x = 2sqroot(22)

see image.

3 0
2 years ago
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
What is the measure of angle z in this figure?
astraxan [27]
Hello!

I've attached a photo for reference.

Lines A and B form straight angles, which  measure 180 degrees. That means that - 

m∠x + m∠y = 180°
m∠y + m∠z = 180°
m∠z + 43° = 180°
43° + m∠x = 180°

Since you're trying to find z, use the solvable equation with z in it:

m∠z + 43° = 180°
180 = z + 43
137 = z

Answer:
m∠z = 137°

3 0
3 years ago
My friends last question pls answer ASAP
Dafna11 [192]

Answer:

64

Step-by-step explanation:

just count the blocks!

6 0
2 years ago
Read 2 more answers
5 • 3n algebra functions
jok3333 [9.3K]
5 x 3n = 15n

^ This is your answer to your question ^
6 0
3 years ago
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