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2 years ago

9

1which relation is a real function ?

1 answer:

Gre4nikov [31]2 years ago

4 0

Hello,

A and C are functions.
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PLEASE SOLVE IT WITH EXPLANATION!!<br>DONT TAKE POINTS :)<br>

zloy xaker [14]

Step-by-step explanation:

We are asked to simply (2√5 + 3√2)². Using formula: (a + b)² = a² + b² + 2ab. Let's say 2√5 = a, 3√2 = b. So,

→ (a + b)² = a² + b² + 2ab

→ (2√5 + 3√2)² = (2√5)² + (3√2)² + 2(2√5)(3√2)

We are aware about the fact that root means 1/2 and square of root means 2/2 that is 1. Using this we get:

→ (2√5 + 3√2)² = 4(5) + 9(2) + 2(2√5)(3√2)

Solve the brackets, to do so first put the like terms in one box.

→ (2√5 + 3√2)² = 4(5) + 9(2) + 2(2*3)(√5)(√2)

Solve the rest calculations.

→ (2√5 + 3√2)² = 20 + 18 + 2(6)(√10)

→ (2√5 + 3√2)² = 38 + 12√10

Option (a) (38 + 12√10) is the correct option.

4 0

2 years ago

Read 2 more answers

What value of n makes the equation true?

dybincka [34]

1(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20 i say the answer 1 or 10

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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(2x^9y^n) (4x²y^10)-8x^11 y^20

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4 0

1 year ago

For the function G defined by G(x)=5x+3, find G(r+5).

ycow [4]

Answer:

8x+5

Step-by-step explanation:

any letter will mean 1

4 0

2 years ago

Read 2 more answers

I need some help please

Veronika [31]

The definition of the tangent function tells you
tan(angle) = (300 ft) / (distance to mountain)
This equation can be rearranged to
(distance to mountain) = (300 ft) / tan(angle)

For the far end of the river,
distance to far end = (300 ft) / tan(24°) ≈ 673.8 ft

For the near end of the river
distance to near end = (300 ft) / tan(40°) ≈ 357.5 ft

Then the width of the river can be calculated by finding the difference of these distances:
width of river = distance to far end - distance to near end
width of river = 673.8 ft - 357.5 ft
width of river = 316.3 ft

The appropriate answer choice is
316 ft.

8 0

3 years ago

What is m in this question

solmaris [256]

-16 because you need to subtract -4 from -12

7 0

3 years ago