All categories

3 years ago

SOMEONE PLEASE HELP ME WITH THIS!!!!!!

Mathematics

1 answer:

Elis [28]3 years ago

7 0

B
Because if you solved it
10z + 20 = 80
-20 -20
10z = 60
/10 /10

You might be interested in

Oliver didn't have any money to play a game at the fair. He borrowed four dollars from Annie and told her he'd pay her next week

Oksi-84 [34.3K]

Answer:

He practicably stole her money

Step-by-step explanation:

that scu m bag

3 0

4 years ago

Read 2 more answers

I WILL MARK BRAINLIEST NEED ASAP PLS

Lelechka [254]

Answer:

option D or y = 7x + 2 is the answer

8 0

3 years ago

How do you solve perimeter of polygons?

earnstyle [38]

<span>idk if this exactly correct but i tried

1 Find and add the lengths of all the polygon's sides
2.Multiply the lengths of equal sides by the number of equal sides.
3 Multiply a regular polygon's side length by the number of sides.
4 Alternatively, use the area and apothegm of the polygon.
</span>

5 0

3 years ago

A street magician asks a passerby to draw one card from a standard deck, remember it, and then replace it. The magician then shu

Romashka-Z-Leto [24]

Answer:

P( At least one of the two cards was a heart)=7/16

Step-by-step explanation:

The solution to this question can be found in two ways.

First way: Since total number of heart in a deck are four,

Probability that neither of two cards had a heart is that at both times the card drawn was not a heart, since it is an independent event, thus probability both were not heart= $\frac{3}{4}$ × $\frac{3}{4}$

= $\frac{9}{16}$

Thus, Probability that at least one of the two cards was a heart= 1- P( Neither of two cards had a heart= 1- $\frac{9}{16}$

= $\frac{7}{16}$

Second way:

Probability that first card drawn is a heart and the second one is not a heart= $\frac{1}{4}$ × $\frac{3}{4}$

= $\frac{3}{16}$

Probability that the first card drawn is not a heart and the second one is a heart= $\frac{3}{4}$ × $\frac{1}{4}$

= $\frac{3}{16}$

Probability that both cards drawn are hearts= $\frac{1}{4}$ × $\frac{1}{4}$

= $\frac{1}{16}$

Adding all these probabilities, we have the probability that at least one of the card drawn was heart= $\frac{7}{16}$

3 0

4 years ago

Simplify -4x^3 x 2y^-2 x 5y^5 x x^-8

n200080 [17]

$-4x^3\cdot \:2y^{-2}\cdot \:5y^5x^{-8}$

$=2\cdot \:4\cdot \:5\cdot \dfrac{1}{y^2}x^{-8}x^3y^5$

$=2\cdot \:4\cdot \:5\cdot \dfrac{1}{x^8}\cdot \frac{1}{y^2}x^3y^5$

$=-\dfrac{1\cdot \:1\cdot \:4x^3\cdot \:2\cdot \:5y^5}{y^2x^8}$

$=-\dfrac{2\cdot \:4\cdot \:5x^3y^5}{x^8y^2}$

$=-\dfrac{40x^3y^5}{x^8y^2}$

$=\dfrac{40y^5}{x^5y^2}$

$=-\dfrac{40y^3}{x^5}$

8 0

4 years ago