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3 years ago

I NEED HELP I HAVE BEEN DOING THIS FOR HOURS I ACTUALLY HAD A MENTAL BREAKDOWN PLEASE HELP ME I WILL MARK BRAINLIEST

Mathematics

2 answers:

PIT_PIT [208]3 years ago

6 0

Answer:

x=8

Step-by-step explanation

Vertical angles are always congruent which means they are equal

Since we know this we can simply set the two angle equal to each other so...

5x-5=2x+19

Now what we do is add five to 19 and subtract 2x from 5x to get

3x=24

simply divide 24 by 3 to isolate x and now we have

x=8

Hope this was helpful!

HACTEHA [7]3 years ago

3 0

If both of them are vertical angles, they equal each other. 2x+19=5x-5 is how you would set it up. Add 5 to 19 to get 24 and subtract 2x from 5x to get 3x. 3x=24, so x=8. If you plug that in, your angles should be 35 degrees for each.

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An office store pays a whole price of $120 for a laser printer and sells it for $174. What is the pecent mark up

Advocard [28]

Answer:

its by 54%

Step-by-step explanation:

because it is raising the price by making money by 54%. and that its using the calculator in any way.

8 0

2 years ago

Solve using the quadratic equation -4x² + 7x + 5 = 0

Luda [366]

Answer:

x=-0.544727

x=2.294727

Step-by-step explanation:

Given quadratic equation to solve:

-4x²+7x+5=0

To solve, use the quadratic formula

x= -b ± b² - 4ac / 2a

Substitute your values for a, b, & c and solve

x= -7 ± 7² - 4·(-4)·5 / 2·(-4)

x= -7 ± 49 - -80/-8

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Then, reduce the problem

x= -7/-8 ± 129/-8

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x=2.294727

4 0

2 years ago

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$