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3 years ago

Which equation describes the relationship between the values in the table?

Mathematics

1 answer:

Sauron [17]3 years ago

7 0

Answer: i think option 2

Step-by-step explanation: sorry if i get it wrong i try my best

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Please help I appreciate your help

steposvetlana [31]

It is twice as the distance hope this helps!!

3 0

3 years ago

Read 2 more answers

I'm really bad at age chart problems. Can someone help me with this and provide an explanation?

Fed [463]

Answer: 10

Step-by-step explanation:

tony is 5 years younger than Marvin's age 1 year from now.

t = (m+1) - 5

t = m - 4

6 years ago tony was half as old as marvin was then.

t - 6 = 1/2(m - 6)

Get rid of the fraction, multiply both sides by 2

2(t-6) = m - 6

2t - 12 = m - 6

2t = m - 6 + 12

2t = m + 6

from the first statement, replace t with (m-4)

2(m-4) = m + 6

2m - 8 = m + 6

2m - m = 6 + 8

m = 14 is Marvin's present age

how old is tony now?

t = 14 - 4

t = 10 yrs is tony's age

4 0

3 years ago

Find the equation of the circle that satisfies the given conditions. radius 4 and center (3, -1)?

dmitriy555 [2]

the equation of this circle is (x - 3)² + (y + 1)² = 4²

5 0

3 years ago

Which of the following is equal to the product of 32 and 1/4 ?

Oxana [17]

Hey there!

The answer to your question is 8

In math, the "product of" represents multiplication, so we multiply the 2 numbers.

32 x 1/4 = 8

Hope this helps! Good luck, and have a great day!

8 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago