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3 years ago

Solve for k 5 - vk = k show work

Mathematics

1 answer:

MissTica3 years ago

3 0

Answer:

The answer to your question is k = $\frac{5}{1 + v}$

Step-by-step explanation:

Equation

5 - vk = k

Process

1.- Add vk in both sides

5 - vk + vk = k + vk

2.- Simplification

5 = k + vk

3.- Factor k (like terms)

5 = k(1 + v)

4.- Divide both sides by "1 + v"

5 / (1 + v) = k(1 + v) / (1 + v)

5.- Simplify

5 / (1 + v) = K

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2z -5(z +2) = -8 - 2z<br><br> Please explain!! Thank you

Korolek [52]

Follow PEMDAS.

You would first do your distributive property which is, -5(z+2). You would multiply -5 to z and -5 to 2 which will give you -5z + -10.

so now you have 2z-5z+ -10= -8 - 2z. Now you must isolate the variable. So subtract -8 to -10 which will give you -18.

Now you have the equation 2z-5z-18=-2z

Now combine like terms. 2z-5z= -3z

Now you have the equation -3z-18=-2z

Now you must move the -3z to the other side of the equal sign by adding the opposite which is +3z to -2z which will give you +1z

Now you have -18= 1z

Now you divide. -18 divided by 1 is -18.

So your final answer will be -18=z

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3 years ago

Please help me complete this !! 12points

Svet_ta [14]

Answer:

x1=1

x2=3

y1=5

y2=1

midpoint= (x1+x2/2, y1+y2/2)=(1+3/2,5+1/2)=(2,3)

x1=2

x2=4

y1=4

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(2+4/2,4+6/2)=(3,5)

x1= -2

x2=4

y1=4

y2= -6

midpoint= (x1+x2/2, y1+y2/2)=(-2+4/2,4-6/2)=(1,-1)

x1=0

x2=0

y1=3

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(0/2,3+5/2)=(0,4)

x1= -2

x2=2

y1= -6

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(-2+2/2,-6+6/2)=(0,0)

x1=1

x2=4

y1=4

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(1+4/2,4+5/2)=(5/2,9/2)

x1=-3

x2=-4

y1=5

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(-3-4/2,5+5/2)=(-7/2,5)

x1=5

x2= -4

y1=2

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(5-4/2,2+6/2)=(1/2,4)

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