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3 years ago

10

Can someone help me solve y 12 + y = 4x

1 answer:

Leviafan [203]3 years ago

7 0

Answer: Y = 4x - 12

Step-by-step explanation:

Well, the reason for this is because y can be anything because there are 2 unknown terms in the equation. Thus, the answer to y can be many different answers. The one common thing that all of these answers have is that they have only 1 expression (An equation without equal sign) to solve it. This expression is the answer to y, but it can be changed judging by the answer of x. That is why you cannot determine an exact answer to y with there being many, so you have to write the expression to get y. Which you can find out by:

12 (Part) + y (Part) = 4x (Whole)

So to figure out one part do the whole minus the other part which is why 4x is our answer:

4x (Whole) - 12 (Other Part)

This is called part-whole postulate.

Hope this helps!

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An equation is given below.

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Answer:

The correct option is D.Any real number value of x satisfies the equation.

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3 years ago

Thirty-six divided by six

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2 years ago

1) Where is the midpoint of , when A (6, 7) and B (-9, -5)?

lord [1]

Answer:

(−1.5,1)

Step-by-step explanation:

Finding the distance, midpoint, slope, equation and the x y-intercepts of a line passing between the two points p1 (6,7) and p2 (-9,-5)

The distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √ ((X2-X1)2+(Y2-Y1)2)

d = √ (-9-6)2+(-5-7)2

d = √ ((-15)2+(-12)2)

d = √ (225+144)

d = √ 369

The distance between the points is 19.2093727122985

The midpoint of two points is given by the formula

Midpoint= ((X1+X2)/2,(Y1+Y2)/2)

Find the x value of the midpoint

Xm=(X1+X2)/2

Xm=(6+-9)/2=-1.5

Find the Y value of the midpoint

Ym=(Y1+Y2)/2

Ym=(7+-5)/2=1

The midpoint is: (-1.5,1)

6 0

3 years ago

Read 2 more answers

Make z the subject 9d= 6z - 5d

yarga [219]

Answer:

$z=2\frac{1}{3}d$

Step-by-step explanation:

$9d=6z-5d$

$6z=9d+5d$

$6z=14d$ (combine like terms)

$z=2\frac{1}{3}d$

Hope this helps :)

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2 years ago

Numerical value of cosh(ln5)

strojnjashka [21]

$\cosh x=\dfrac{e^x+e^{-x}}2$

So

$\cosh(\ln 5)=\dfrac{e^{\ln5}+e^{-\ln5}}2=\dfrac{5+\frac15}2=\dfrac{13}5$

6 0

3 years ago