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Setler [38]
3 years ago
11

Eat brainlyest cheese

Mathematics
2 answers:
weeeeeb [17]3 years ago
5 0

Answer:

Oooo cheese- wuts ur favorite type of cheese? :3

Step-by-step explanation:

:3

umka2103 [35]3 years ago
4 0

Answer:

fdiyshufsdhbdjfhsdfushdifusfhdifuhsd

Step-by-step explanation:

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What's is 7x90 it not hard it's not easy it medium
Leya [2.2K]
90 × 7 = 630 , it's not hard, it's easy
8 0
3 years ago
You have 5 Red Marbles, 4 Green Marbles, and 3 Blue Marbles in a Bowl. You randomly pick 4 marbles from the bowl (without replac
makvit [3.9K]

Answer:

2%

Step-by-step explanation:

In this case each one is an independent event, therefore, the multiplication of each one would be the final probability

We have 5 Red Marbles, 4 Green Marbles, and 3 Blue Marbles, that is, there are 5 + 4 + 3 12 Marbles in total.

Now if I draw a red one, the probability would be: 5/12

When drawing another red, the probability would be: 4/11

When taking the green: 4/10

When removing the blue: 3/9

Finally, the final probability is:

P = (5/12) * (4/11) * (4/10) * (3/9)

P = 0.020

In other words, the probability of this happening is 2%

7 0
3 years ago
A (p + 1) metres long ladder reaches a height
statuscvo [17]

p^2 + (2p - 5)^2 = (p + 1)^2

Solve for p to find your answer.

4 0
3 years ago
X/5+7=3<br> tellme fastplease
Aliun [14]

Answer:

-20

Step-by-step explanation:

x/5=3-7

x/5=-4

x=-4×5

x=-20

5 0
2 years ago
Help, please! If possible explain it like your talking to a little kid :) Thank you♡
monitta

Answer:

GCF: y³z³

Step-by-step explanation:

The greatest common factor is the a term that you can take out of all the terms given to you. This term, when multiplied to each of the individual numbers in the set, will return you to the original amount.

In this case, note that they all share the variables y and z, and that each of them have <em>at least</em> 3 y's and 3 z's. For you to factor, you will divide these from all the terms.

\frac{x^{3}y^{5}z^{5} , y^{3}z^{5} , xy^{3}z^{3} }{y^{3}z^{3}} = x^{3}y^{2}z^{2} , z^{2} , x

y³z³(x³y²z² , z² , x)

~

7 0
2 years ago
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