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4 years ago

2x-y=5solve the equation for y

Mathematics

1 answer:

Taya2010 [7]4 years ago

7 0

2x-y=5 Subtract 2x from both sides -y=-2x+5 Multiply everything by -1 y=2x-5 would be your answer

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There are 42 students 21 are in 7th grade 14 are in 8th 7 are in ninth whats the probability of an eighth grader winning

Varvara68 [4.7K]

Answer:

The probability of an eighth grader winning is 1/3

Step-by-step explanation:

Total no. of students = 42

Number of students in 7th grade = 21

Number of students in 8th grade = 14

Number of students in 9th grade = 7

Probability of event = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So,the probability of an eighth grader winning= $\frac{14}{42}=\frac{2}{6}=\frac{1}{3}$

Hence The probability of an eighth grader winning is 1/3

3 0

3 years ago

Help I don't get it can u explain it to me

kap26 [50]

You are given the masses of four planets.

Add them together, and then divide your sum by 4.

That will be the "average planet mass" for these 4 planets.

3 0

3 years ago

Question 14 of 25

Naily [24]

Answer:

Step-by-step explanation:

perpendicular lines are the negative reciprocals of each other, and a pair of these lines intersects at 90 degrees.

5 0

3 years ago

Read 2 more answers

What is 112.491 written in word form?

bearhunter [10]

Answer:

112.491 = one hundred twelve and four hundred ninety-one

thousandths.

Step-by-step explanation:

6 0

3 years ago

7.<br>(-3x³)(-2x³y⁴z)(-3z²)÷(4x³z)(-3yz) (-3xyz)

serg [7]

Answer:

$-\frac{x^2y^2}{2}$

Step-by-step explanation:

We have an extremely large equation and are asked to divide it, so let's solve it step-by-step :

Remove the parenthesis to make it easier to read :

$\frac{-3x^3 *2x^3y^4z *3z^2}{4x^3z*3yz*3xyz}$

Multiply the numerators :

$\frac{-18x^6y^4z^3}{4*3*3x^3yyzzz}$

Multiply the denominators :

$\frac{-18x^6y^4z^3}{36x^4y^2z^3}$

Apply the negative rule :

$-\frac{18x^6y^4z^3}{36x^4y^2z^3}$

Cancel the common factor which is 18 :

$-\frac{x^6y^4z^3}{2x^4y^2z^3}$

Apply the addition exponent rule :

$\frac{y^4z^3x^{6-4} }{2y^2z^3}$

Subtract :

$\frac{x^2y^4z^3 }{2y^2z^3}$

Apply the rule for y :

$\frac{x^2y^{4-2} z^3 }{2z^3}$

Subtract :

$\frac{x^2y^2z^3 }{2z^3}$

Cancel the common factor of z^3 :

$-\frac{x^2y^2}{2}$

5 0

3 years ago