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2 years ago

Sas, sss, aas, or not enough information

Mathematics

2 answers:

maria [59]2 years ago

8 0

Answer:

Sas

Step-by-step explanation:

Because it’s side angle side

Ksju [112]2 years ago

7 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The given triangles are congruent by :

SAS criterion

because two of their sides are equal and the angle between the is equal too (by vertical opposite angle pair).

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Area and perimeter to find the length of it

dusya [7]

Wht plz tell me wht u need do u want the formula

6 0

2 years ago

Read 2 more answers

13 Simplify the following expression. -84 = 12 O A. 7 oc. o D. 8 o v

padilas [110]

Answer:

435

Step-by-step explanation:

8 0

2 years ago

What is the smallest angle of a right triangle with sides 9, 40, and 41? Round your answer to the nearest degree, and enter the

svlad2 [7]

Answer: 13°

Explanation: x = tan^-1 ( 9/40) ≈ 13°

8 0

2 years ago

DO THE MATH

lesantik [10]

f(p) where p is the price in thousands. f(250) means the average number of days before being sold for $250,000. Lets take a look at the answers and see which fits the best:

1. The house sold for $250,000. (This is what the 250 stands for, but we need to find what f(250) means and not just 250)

2. The house stayed on the market for an average of 250 days before being sold. (Nope, not even close)

3. This is the average number of days the house stayed on the market before being sold for $250,000. (Yes! This seems right, f(250) is the average number of days before being sold for $250,000)

4. The house sold on the market for $250,000 and stayed on the market for an average of 250 days before being sold. (Nope, we are not told anywhere that it takes 250 days to be sold)

This means that our answer has to be 3) This is the average number of days the house stayed on the market before being sold for $250,000.

I hope I've helped! :)

4 0

2 years ago

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^

Sedbober [7]

Hello,

a)
$I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\

= [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\

=0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\

= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\
= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\

$
$I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
$

b)
$\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\
= \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\
= \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\




$

$\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\
= \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}\\





$

c)

$I_n= \dfrac{n-1}{n} * I_{n-2} \\

I_{2n+1}= \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\
= \dfrac{2n}{2n+1} * I_{2n-1} \\
= \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\
= \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\

I_1=1\\

$

3 0

3 years ago