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otez555 [7]
2 years ago
10

Please help the best answer I'll give brainliest

Mathematics
1 answer:
ad-work [718]2 years ago
8 0
X intercept is 5, 13 y is 5, -13
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Maurice puts 130 trading cards in projector sheet he feels 7 sheets and puts the remaining four cards in an 8 sheets have the sa
cestrela7 [59]

Step-by-step explanation:

i am expecting that total cards is 234

7 0
3 years ago
PLEASEEEE HELPPPPPP!!!
artcher [175]

Answer:

18

Step-by-step explanation:

3 0
3 years ago
Which inequality represents the statement? The ages (a) of s group of travelers all exceeded 17 years and were no more then 54 y
klasskru [66]

Answer:

The inequality that represents the age of the group, "x", is: 17 < x \leq 54

Step-by-step explanation:

To express this problem in an inequality we will attribute the age of the members on the group with the variable "x". There are two available information about "x", the first states that every member of the group is older than 17 years, therefore we can create a inequality based on that:

x > 17

While the second data from the problem states that none of than is older than 54 years old, this implies that they can be at most that old, therefore the inequality that represents this is:

x\leq 54

In order for both to be valid at the same time x must be greater than 17 and less or equal to 54, therefore we finally have:

17 < x \leq 54

7 0
3 years ago
The velocity-time graph for a cycle is shown
malfutka [58]

Answer:

<u>a) 100m</u>

<u>b) 2m/s</u>

Step-by-step explanation:

a)==>> <u>2+4</u>+<u>6+8</u>+10(8) = 100m

b)==>> Variable "a" = change OVER time (\frac{change}{time}) = 8m/4s = 2m/s

5 0
3 years ago
(1+cos2x)/(1-cos2x) = cot^2x
sesenic [268]

We will turn the left side into the right side.

\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x

Use the identity:

\cos 2x = \cos^2 x - \sin^2 x

\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x

\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x

Now use the identity

\sin^2 x + \cos^2 x = 1 solved for sin^2 x and for cos^2 x.

\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x

\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x

\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x

\cot^2 x = \cot^2 x


8 0
3 years ago
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