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3 years ago

PLZ ANSWER PLZ PLZ PLZ HELP 1 QUESTION!!!

Mathematics

1 answer:

JulsSmile [24]3 years ago

8 0

We first must find the value of angle z.

Angle z = 90° - 42°

Angle z = 48°

We now use the sine function to find y.

sin(48°) = 35/y

y = 35/sin(48°)

y = 47.0971455362

We now round off to two decimal places.

y = 47.10

I hope this helps....

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Ayonna has $300 gift card. She buys video games that cost $50 each . After

abruzzese [7]

Answer:

Step-by-step explanation:

300- x =100

50+50=100

100+100=200

300-200=100

so she bought 4 video games

5 0

3 years ago

Can someone plzzz hekppp meeee

erastova [34]

Answer:

Use this formula for these kinda things.

Step-by-step explanation:

Remember The z is the hypathonous, it means the biggest side.

${a}^{2} + {b}^{2} = {c}^{2}$

where Z is going to be C

put this all together

${7}^{2} + {18}^{2} = {z}^{2}$

put it into calculator, means 373=

${z}^{2}$

therfore take the square root, it will be

$z = \sqrt{373}$

which it will be 19.31

5 0

3 years ago

Caroline is considering two video game rental plans. Plan A can be modeled with the equation C = 2n, and Plan B can be modeled w

meriva

Options

A. Caroline rents exactly 7 games each month.

B. Caroline rents exactly 6 games each month.

C. Caroline rents 6 or more games each month.

D. Caroline rents from 1 to 5 games each month.

Answer:

D. Caroline rents from 1 to 5 games each month.

Step-by-step explanation:

Given

Plan A:

$C = 2n$

Plan B:

$C = n + 6$

Required

Which options justifies A over B

The solution to this question is option (d).

In option d, n = 1,2,3,4,5

When any of the values of n is substituted in plan A and B, respectively; the cost of plan A is cheaper than plan B.

This is not so, for other options (A - C)

To show:

Substitute 1 for n in A and B

Plan A:

$C = 2n$ $= 2 * 1 = 2$

Plan B:

$C = n + 6$ $= 1 + 6 = 7$

Substitute 5 for n in A and B

Plan A:

$C = 2n$ $= 2 * 5 = 10$

Plan B:

$C = n + 6$ $= 5 + 6 = 11$

8 0

3 years ago

Help me.................

-Dominant- [34]

Sry but I need more information to answer the question. So so sry!

6 0

3 years ago

Read 2 more answers

a standard deck of cards with 26 red cards and 26 black cards is split into two piles, each having at least one card. In pile A

Fantom [35]

Answer: There are a total of 22 CARDS in pile B

Step-by-step explanation:

It is already certain that the total number of cards that are in pile "A" must be a multiple of 5 due to the fact that there are four times as many black cards as red cards.

If there are 26 red cards and 26 black cards altogether in the piles and each pile must have at least one card, then let's look at the possibilities:

4 black cards, 1 red card in pile A, it will then result to 22 black cards and 25 red cards in pile B. Recall that the number of red cards in pile B must be a multiple of the number of cards in pile B

We check. 25/22 .... (Doesn't agree to multiple rule)

We try another way, 8 black cards, 2 red cards in pile A which then results to 18 black cards and 24 red cards in pile B

(24/18) doesn't agree.

Again 12 black cards, 3 red cards in A, 14 black cards and 23 red cards in pile B... doesn't agree

16 black cards, 4 red cards in pile A, 10 black cards and 22 red cards in pile B... doesn't agree

20 black cards, 5 red cards in pile A, 6 black cards and 21 red cards in pile B... doesn't agree

24 black cards, 6 red cards in pile A, 2 black cards and 20 red cards in pile B... We check if the number of red cards in pile B is a multiple of the number of black cards in pile B:

20/2 = 10 (it agrees to the multiple rule).

So the number of cards in pile B = (number of red cards there + number of black cards there)

= 20 + 2

= 22 cards are in pile B altogether

4 0

3 years ago