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3 years ago

Can you find the value of y

Mathematics

1 answer:

SashulF [63]3 years ago

8 0

Answer:

y = 18

Step-by-step explanation:

The angles 8y - 102 and 2y + 6 are vertical and congruent, thus

8y - 102 = 2y + 6 ( subtract 2y from both sides )

6y - 102 = 6 ( add 102 to both sides )

6y = 108 ( divide both sides by 6 )

y = 18

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3 years ago

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In the following figure it is given that 2a-5b=10 find a and b

RoseWind [281]

Answer:a=130,b=50

Step-by-step explanation:

a+b= 180

2a-5b=10

2(180-b)-5b=10 (I have substituted (180-b) for a here.)

360-2b-5b=10

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3 years ago

the half-life of strontium-90 is approximately 29 years. how much of a 500 g sample of strontium-90 will remain after 58 years

Yuliya22 [10]

Answer: 125 g

<u>Step-by-step explanation:</u>

$A = P_o\cdot e^{kt}\\\\\text{First, use the given information to find k:}\\\\\bullet A=\dfrac{1}{2}P_o\\\\\bullet k = unknown\\\\\bullet t=29\text{ years}\\\\\dfrac{1}{2}P_o=P_o\cdot e^{k(29)}\\\\\\\dfrac{1}{2}=e^{k(29)}\qquad divided\ both\ sides\ by\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=ln\bigg(e^{k(29)}\bigg)\qquad applied\ ln\ to\ both\ sides\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=29k\qquad simplified-ln\ and\ e\ cancel\ out\\\\\\\dfrac{ln\bigg(\dfrac{1}{2}\bigg)}{29}=k\qquad divided\ 29\ from\ both\ sides\\\\\\-0.0239=k$

$\text{Now, use the following in the equation to solve for A:}\\\\\bullet A=unknown\\\bullet P_o=500\\\bullet k=-0.0239\\\bullet t=58\text{ years}\\\\A=500\cdot e^{(-0.0239)(58)}\\\\.\quad=500\cdot e^{-1.386}\\\\.\quad=125$

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4 years ago

A stadium has a a central rectangular area 125m long by 80m wide.there are two semi circular ends.a running track goes all the w

Fudgin [204]

a)The total inside length of the curved ends of the track will be 25.13 meters

b)The total distance around the inside of the track will be 275.13 meters

c)The area inside the track is 10050 square meters.

The space occupied by any two-dimensional figure in a plane is called the area. The space occupied by the circle in a two-dimensional plane is called the area of the circle.

Given that:-

A stadium has a central rectangular area 125m long by 80m wide. there are two semi-circular ends. a running track goes all the way around.

a) what is the total inside length of the curved ends of the track?

The length of the curved track means it is a perimeter of the circle so the radius of the semicircle will be 4 meters.

So the length will be = π r + πr = 2 πr

= 2 x π x 4 = 25.13

b) what is the total distance around the inside of the track?

The total distance of the track will be the sum of the length of the two semicircles and the longer side of the rectangle.

Total length = 125 + 125 + 25.13

= 275.13 meters

c) what is the area inside of the track?

The area will be equal to the area of the two semicircular tracks and the rectangular track.

Total Area = Area of semicircle + Area of semicircle + Area of rectangle

= ( π/2 ) r² + (π/2) r² + ( L x W )

= π r² + ( L x W)

= π ( 4 )² + ( 125 x 8 )

= 10050.26 square meters.

Therefore the total inside length of the curved ends of the track will be 25.13 meters. The total distance around the inside of the track will be 275.13 meters. The area inside the track is 10050 square meters.

To know more about an area follow

brainly.com/question/25292087

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