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3 years ago

Brendan ordered some

Mathematics

1 answer:

tamaranim1 [39]3 years ago

6 0

Answer:

he bought 4 games

Step-by-step explanation:

since each game is 22$ if you bought 4, the amount would equal 88 and plus 9$ for shipping you add it all together and you get 96$

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Helppp and explain plzzzzzz

devlian [24]

The second one is correct. 66/300 = x/100

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4 years ago

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6) Which is an equation of the line with a slope of -4 and a y-intercept of -2?

DedPeter [7]

Answer:

its 4x+y-2 I think but I am not totally sure

Step-by-step explanation:

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%7C%5Cfrac%7Bx%2B1%7D%7Bx-1%7D%2B1%7C%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%7Bx%2B1%7D%7Bx-1%

VMariaS [17]

The inequality boils down to

|y| > y

By definition of absolute value, we have

• |y| = y if y ≥ 0

• |y| = -y if y < 0

So if y ≥ 0, we have

y > y

but this is a contradiction.

On the other hand, if y < 0, we have

-y > y ==> 2y < 0 ==> y < 0

and no contradiction.

Now replace y with (x + 1)/(x - 1) + 1. Then you're left with solving

(x + 1)/(x - 1) + 1 < 0

(x + 1 + x - 1)/(x - 1) < 0

2x/(x - 1) < 0

The left side is negative if either 2x > 0 and x - 1 < 0, or 2x < 0 and x - 1 > 0. The first case reduces to x > 0 and x < 1, or 0 < x < 1. In the second case, we get x < 0 and x > 1, but x cannot satisfy both conditions, so we throw this case out.

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3 years ago

In how many ways can 2 red, 2 black, 3 white and 2 blue balls be selected from 4 red, 3 black, 4 white and 8 blue balls? In how

Montano1993 [528]

From the total pool of colored balls, one can choose 2 reds, 2 blacks, 3 whites, and 2 blues in

$\dbinom42\cdot\dbinom32\cdot\dbinom43\cdot\dbinom82=6\cdot3\cdot4\cdot28=2016$

ways.

I'm assuming no ball of the same color is distinguishable from any other ball of the same color. So when I'm considering the possible arrangements, if I had lined up the ball as

red1 - black - red2 - ...

then this would be no different that

red2 - black - red1 - ...

So I now have 9 balls to arrange, which means there are $9!=362,880$ total possible permutations of them. But order among distinct colors is assumed to not matter. This means I have to divide the total number of permutations by the number of ways I could permute balls of the same color. Then there would be a total of

$\dfrac{9!}{2!\cdot2!\cdot3!\cdot2!}=7,560$

ways of arranging the balls I had selected.

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3 years ago

How to solve a proportion

tatuchka [14]

Answer:

Step-by-step explanation:

$\frac{x}{5}$ · $\frac{12}{15}$ =

15x=12 · 5

15x = 60

x = 4

7 0

3 years ago

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