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2 years ago

How do I find the value of each variable?

Mathematics

1 answer:

arlik [135]2 years ago

7 0

You have to solve for x and plug it back in

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Does anyone know the answers to this algebra question? Will give brainiest

MaRussiya [10]

Option 1 : -22

using formula

(a + b )(a - b ) = a^2 - b^2

(√10 + 2√8) (√10 - 2√8)

= (√10)^2 - (2√8)^2

= 10 - 2×2×8=10-32

= -22

2nd question answer is option 3

3 0

2 years ago

Brianna has a sheet of paper that is 6 feet long.she cuts the length of paper into sixth and then cuts the length of each of the

Inga [223]

72 inches i think im not sure

7 0

2 years ago

Could someone please help me with #2? Thank you :)

iren [92.7K]

Answer:

The answer is B.

Step-by-step explanation:

4 is a multiple of 2, which explains the reasoning of answer choice B.

Answer choice A is wrong because 2 is Not a multiple of 4.

Answer choices C and D are incorrect because it takes multiple steps to solve for x and substitute into the other equation. They claim that it takes on step to solve for x, but it would take 2 steps to solve for x.

#teamtrees #WAP (Water And Plant)

3 0

2 years ago

PLZ HELP NEED ASAP, much appreciated

Margaret [11]

Answer:

Ok that's not it

x=-1 then y=4

x=0 then y= 6

x=1 then y = 8

x=2 then y= 10

x=3 then y= 12

Step-by-step explanation:

when y=2x+6

you have to plug what it says x is in each of the equations.

For the first one it says x=-1

you then plug that in and you get y=2(-1)+6

then you go through PEMDAS and first you get y=-2+6

-2+6 = 4

therefore y=4

and you just go through the rest wiht that

Step-by-step explanation:

6 0

1 year ago

If two standard six-sided dice are tossed, what is the probability that a 5 is rolled on at least one of the two dice? express y

zaharov [31]

We define the probability of a particular event occurring as:
$\frac{number\ of \ desired\ outcomes}{number\ of\ possible\ outcomes}$

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are independent, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for each of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.

Now that we've found the number of possible outcomes, we need to find the number of desired outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is at least one 5 rolled. It turns out, there are only 3:

(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5

So, we have $\frac{3}{36} = \frac{1}{12}$ probability of rolling at least one 5.

6 0

2 years ago