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3 years ago

Can you conclude that these triangles are congruent? Yes or no

Mathematics

2 answers:

vfiekz [6]3 years ago

7 0

Answer:

Yes

Step-by-step

If there flipped then they’re congruent

vitfil [10]3 years ago

5 0

Answer:

Yes

Step-by-step explanation:

Here we get two triangles,

In triangle MNH. In triangle JKH

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A chemical spill is leaking into a river flowing 2 3/4 miles per hour. How long will it take the chemicals to reach a town locat

ExtremeBDS [4]

Answer:

$6\frac{3}{11}=6.27$ hours.

Step-by-step explanation:

We have been given that a chemical spill is leaking into a river flowing 2 3/4 miles per hour.

To find the time taken by chemical to reach the town we will divide distance of town from the spill by distance covered by chemical spill in one hour.

$\text{Time taken by chemical to reach the town}=17\frac{1}{4} \div 2\frac{3}{4}$

Upon converting our mixed fractions into improper fractions we will get,

$\text{Time taken by chemical to reach the town}=\frac{69}{4} \div \frac{11}{4}$

Dividing a fraction with another fraction is same as multiplying first fraction by the reciprocal of second fraction.

$\text{Time taken by chemical to reach the town}=\frac{69}{4}\times \frac{4}{11}$

After cancelling out 4 we will get,

$\text{Time taken by chemical to reach the town}=\frac{69}{11}$

$\text{Time taken by chemical to reach the town}=6\frac{3}{11}$

$\text{Time taken by chemical to reach the town}=6.272727\approx 6.27$

Therefore, it will take the chemicals $6\frac{3}{11}=6.27$ hours to reach the town.

5 0

4 years ago

1. "The sum of a number and 4.

Norma-Jean [14]

Answer:

a+4=

Step-by-step explanation:

7 0

3 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

4 years ago

<img src="https://tex.z-dn.net/?f=%28%20-%20%2055%292" id="TexFormula1" title="( - 55)2" alt="( - 55)2" align="absmiddle" clas

Art [367]

The answer is -110.
This is because -55-55=-110.
You have to put -55-55 because the question is asking for you to multiply -55 twice, which is technically-55-55.

8 0

3 years ago

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How does buying items in bulk usually help a consumer?

cluponka [151]

You get more stuff for a better price.....?( Idk)

4 0

3 years ago

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