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1 year ago

Find the values of X and Y

Mathematics

1 answer:

Vitek1552 [10]1 year ago

7 0

Answer:

x=90, angle bisector theorem

y=43,

90+47+y=180

=137+y=180

y=43

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Please find out the answer and I will mark your answer as the brain test with a five-star rating and a thank you. But only if th

pentagon [3]

Answer:

Below

Step-by-step explanation:

Let x be that missing number

One third of it is x/3

One-ninth of it is x/9

Multiply x/3 and x/9

● (x/3)*(x/9) = (x^2/27)

● (x^2/27) = 108

Multiply both sides by 27

● (x^2/27)*27 = 108*27

● x^2 = 2916

● x = √(2,916) or x = -√(2,916)

● x = 54 or x = -54

So there are two possibilities 54 and -54.

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2 years ago

Read 2 more answers

A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 3 centimeters. Assuming the balloon i

Anni [7]

Answer:

The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.

Step-by-step explanation:

We are given the following information in the question:

A spherical balloon is designed to pop at the instant its radius has reached 3 centimeters.

The balloon is filled with helium at a rate of 10 cubic centimeters per second .

$\displaystyle\frac{dV}{dt} = 10~cm^3/sec.$

We have to find fast the radius is growing at the instant it pops.

Volume of spherical balloons =

$\displaystyle\frac{4}{3}\pi r^3$

Differentiating, we get,

$\displaystyle\frac{dV}{dt} = \frac{4}{3}\pi 3r^2 \displaystyle\frac{dr}{dt} = 4\pi r^2 \displaystyle\frac{dr}{dt}$

Putting the values, we get,

$10 = 4\times 3.14\times (3)^2\times \displaystyle\frac{dr}{dt}\\\\\frac{dr}{dt} = \frac{10}{4\times 3.14\times (3)^2} = 0.0884 \approx 0.09\\\\\frac{dr}{dt} = 0.09\text{ centimeter per second}$

The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.

4 0

2 years ago

HELP PLEASE AND THANK YOU 4JAAB

Maslowich

Answer:

$\bold{Q13}\ \angle W\\\\\bold{Q14}\ \angle S\\\\\bold{Q15}\ side\ WA\\\\\bold{Q16}\ \triangle PTA$

Step-by-step explanation:

For Q13, Q14, Q15

$\text{for}\ \overline{WA}\ \text{and}\ \overline{AS}\ -\ A\ \text{is common}.\ \text{Therefore inclu}\text{ded angle is}\ \angle A\\\\\text{for}\ \overline{WS}\ \text{and}\ \overline{WA}\to\angle W\\\\\text{for}\ \overline{WS}\ \text{and}\ \overline{SA}\to\angle S\\\\============================$

$\text{Inc}\text{lude side between}\ \angle A\ \text{and}\ \angle W\ \text{is}\ AW\ (WA).\\\\\text{Inc}\text{lude side between}\ \angle S\ \text{and}\ \angle W\ \text{is}\ SW\ (WS).\\\\\text{Inc}\text{lude side between}\ \angle A\ \text{and}\ \angle S\ \text{is}\ AS\ (SA).$