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vladimir2022 [97]
3 years ago
10

How do you simplify this problem?

Mathematics
1 answer:
Yuri [45]3 years ago
4 0

1+tan^2(x) = sec^2(x)

sec^2(x) = 1/cos^2(x)

csc(-x) / 1/cos^2(x) =

cos^2(x) csc(-x)

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Y = 15x + 10, where y is the total cost of renting 1 bicycle on the boardwalk for x hours. Determine whether the function is pro
vova2212 [387]

Answer:

The function is non-proportional.

Step-by-step explanation:

If the function was proportional, when the number of hours spent riding the bicycle is equal to 0, the total cost of renting the bicycle would also be 0. In other words, the function would be proportional if y=0 when x=0. However, this is not the case.

y=15x+10

Plug 0 into the equation as x

y=15(0)+10\\y=10

When 1 bicycle has been rented for 0 hours, the customer still has to pay 10 dollars just for renting it. When x=0, y=10. This means that the function is non-proportional.

I hope this helps!

4 0
3 years ago
When the sum of 8 and a certain number is increased by 12, the result is the same as the product of 3 and the number. What is th
Roman55 [17]
ANSWER: The number is 10.
Hope this helps!!

I used the variable n as the number we needed to find. The equation was n + 8 + 12 = 3n

6 0
3 years ago
Find the measure of a. A. 110 B. 125 C. 55 D. 75
tangare [24]
The answer to your question is c 55
5 0
2 years ago
Use the Distributive Property to multiply 8 and 45 in your head.
Andre45 [30]

Answer:

320+40=360

Step-by-step explanation:

8(40+5)

320+40=360

5 0
3 years ago
Read 2 more answers
PLEASE HELP.
pentagon [3]

Let x represent the number of shirts. Let y represent the number of pens.

If shirts are on sale for $11.99 each, then x shirts cost $11.99x.

If pants are on sale for $12.99 each, then y pants cost $12.99y.

The total cost is $(11.99x+12.99y).

Sarah can spend up to $65. Then an inequality that represents this situation is

11.99x+12.99y≤65 (this inequality holds when Sarah can spend $65 too)

or

11.99x+12.99y<65 (this inequality holds when Sarah can spend less than $65).

3 0
3 years ago
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