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2 years ago

What is the solution to -6x = -24

Mathematics

2 answers:

mars1129 [50]2 years ago

7 0

Answer:

x= 4

Step-by-step explanation:

since x represent a variable. You are supposed to divide both sides by -6 to get that variable. So

-6x=-24/-6

(-24/-6 is 4)

drek231 [11]2 years ago

4 0

X=4. Divide both side by -6 to get x by itself

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myrzilka [38]

Here is the anwser to the problem

6 0

3 years ago

N+7/d+7=3/4 solved for d.

spin [16.1K]

D = -28 / 4N+25
Hope this helps.

8 0

3 years ago

What does 49.39 round to.

Verizon [17]

If you want to round 49.39 to the nearest tenths place you will have to look at your tenths digit. The digit is 3, right? Look at the next number after 3. It is the number 9 for the hundredths place. If the number is bigger than 5 you round it. If it is smaller, you keep it the same. 9 is bigger than 5 so you make the 3 into a 4. Then, you're answer is 49.4 or 49.40 (Which are basically the same answer) It also works the same with the other digits. Simple, right? :)

4 0

3 years ago

F(x)=2x+3/4x+5 find f(-9) pls solve it

Marysya12 [62]

Answer:

15/31

Step-by-step explanation:

substitute -9 were there's x in the function

f(-9)=2(-9)+3/4(-9)+5

=-18+3/-36+5

=-15/-31

the negatives cancel giving you

15/31

4 0

2 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :