MATHEMATICS COLLEGE

What is the answer ?
x/y+17
If is x=12
and y=1/2

Answers

Answer 1
Answer: 12/ 1/2 +17 = 24 +17= 41
Answer 2
Answer:

Answer:

jhhdh.duuemmeyyeymruur*

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Solve used the basic percent equation 10% of 600 is what ??? Remember Percent×base=amount

Answers

Your answer is 60 because you multiply 600 with .10 and get 60.00
10%=0.1
0.1×600=60
10% of 60 is 60

Two sides of a triangle have lengths 9 and 15. What must be true about the length of the third side? less than 6 less than 15 less than 9 less than 24

Answers

Answer:

D. 6<x<24.

Step-by-step explanation:

Let x be the length of third side of our triangle.

We have been given that two sides of a triangle have lengths 9 and 15.

To find the possible length for 3rd side of our triangle we will use triangle inequality theorem.

Triangle inequality theorem states that the sum of the lengths of any two sides of a triangle should be greater than the length of the third side.

Using triangle inequality theorem we will get:

x+9-9>15-9

x>6

x<9+15

x<24

Therefore, the third side of triangle must be greater than 6 and less than 24 and option D is the correct choice.

Determine if the transformation of figure a to figure b is a dilation

Answers

it is a dilation but it's a growth as well, it dilated but shifted as well

30 points!!! Help me on 1. 2. 3.

Answers

I think 1 is to press C

If given f(x)=2x+5 what is f(-10)

Answers

To solve input -10 into x to get your answer, it should look something like this (f(-10) = 2(-10) + 5

I need help plz plz plz help me ASAP. And SHOW YOUR WORK

Answers

r= 8

C= 2,5•r+15
C= 2,5•8+15
C= 20+15
C= 35


Nina's total cost will be 35 $, £, € (i don't know which of them)

Answer:

The cost of the wristband for 8 rides is $35.

Step-by-step explanation:

The expression for the cost, C, for r rides is

C = 2.5r + 15

Since she wants to go on 8 rides, r is 8.

Substitute r with 8 in the cost equation and evaluate it.

C = 2.5r + 15

C = 2.5 * 8 + 15

C = 20 + 15

C = 35

The cost of the wristband for 8 rides is $35.
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