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3 years ago

HELP ME PLS Solve x-3y-6=0

Mathematics

1 answer:

Ne4ueva [31]3 years ago

7 0

Answer: x=3y+6
Explanation to solve:
You must move all terms not containing x to the right side of the equation!

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Answer: 24.2

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3 years ago

Gabriel surveyed 1000 of the students in his school about their favorite color. 55% said their favorite color was purple. How ma

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3 years ago

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X/5 -8=4 find the value of x

elena-s [515]

Answer:

x=60

Step-by-step explanation:

I'm assuming you meant this: (x/5)-8=4

In which case you would add 8 to both sides to get rid of the 8 on the left (your goal is to get x by itself so you want to move the numbers on the x side to the other side of the equal sign)

(x/5)=12

Then you would multiply 5 on both sides to get rid of the fraction with the 5 on the bottom on the left side.

x=60

There's your answer.

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3 years ago

If a number becomes 2.5 times smaller, by what percent was it decreased?

Gwar [14]

Answer: 40%

Step-by-step explanation: 1 Divided by 2.5 is 0.4, or 40%.

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3 years ago

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Complete the steps to solve the polynomial equation x3 – 21x = –20. According to the rational root theorem, which number is a po

jeka94

Answer:

Zeroes : 1, 4 and -5.

Potential roots: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ .

Step-by-step explanation:

The given equation is

$x^3-21x=-20$

It can be written as

$x^3+0x^2-21x+20=0$

Splitting the middle terms, we get

$x^3-x^2+x^2-x-20x+20=0$

$x^2(x-1)+x(x-1)-20(x-1)=0$

$(x-1)(x^2+x-20)=0$

Splitting the middle terms, we get

$(x-1)(x^2+5x-4x-20)=0$

$(x-1)(x(x+5)-4(x+5))=0$

$(x-1)(x+5)(x-4)=0$

Using zero product property, we get

$x-1=0\Rightarrow x=1$

$x-4=0\Rightarrow x=4$

$x+5=0\Rightarrow x=-5$

Therefore, the zeroes of the equation are 1, 4 and -5.

According to rational root theorem, the potential root of the polynomial are

$x=\dfrac{\text{Factor of constant}}{\text{Factor of leading coefficient}}$

Constant = 20

Factors of constant ±1, ±2, ±4, ±5, ±10, ±20.

Leading coefficient= 1

Factors of leading coefficient ±1.

Therefore, the potential root of the polynomial are $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ .

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3 years ago