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4 years ago

Given: RT ≅ TV and ST ≅ TU

Mathematics

2 answers:

den301095 [7]4 years ago

7 0

Answer is - <u>A</u>

( △RTS ≅ △VTU and △RTU ≅ △VTS )

got a 100% on the test - edge 2021

Katarina [22]4 years ago

3 0

Answer:

△RTS ≅ △VTU and △RTU ≅ △VTS should be your answer :)

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Hey guys in the middle of my final! please help? will mark the brainiest + 30 points !

MrRa [10]

Answer:

X=90

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3 0

3 years ago

Based on the given information, what is the measure of the missing length, c?

salantis [7]

Answer:

$c=5\ units$

Step-by-step explanation:

we know that

In the right triangle ABC of the figure

Applying the Pythagoras Theorem

$AC^{2}=AB^{2}+BC^{2}$

substitute the given values

$13^{2}=c^{2}+12^{2}$

Solve for c

$169=c^{2}+144$

$c^{2}=169-144$

$c^{2}=25$

$c=5\ units$

4 0

4 years ago

If four bells toll at intervals of 8,9,12 and 15 minutes respectively. If they toll at 3 pm when will they toll together

miv72 [106K]

Ur basically looking for the common multiple of the numbers 8,9,12,and 15.
LCM of these numbers is 360.

so 360 minutes = (360/60) = 6 hrs

and 6 hrs from 3 p.m. would be : 9 p.m. <== ur answer

7 0

3 years ago

ABCD is a parallelogram the ratio of angle A and Angle B of this Parallelogram is 5 ratio 4 find the measure of Angle B.

sergij07 [2.7K]

Answer:

Mhm lemme see bou this one

Step-by-step explanation:

7 0

3 years ago

Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)

andrezito [222]

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

3 0

4 years ago