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grandymaker [24]
3 years ago
14

8z-7=3z-7 5z how do you solve?

Mathematics
1 answer:
gladu [14]3 years ago
6 0
8z-7=3z-7
-3z -3z

5z-7= -7
+7 +7

*5z=0
You might be interested in
Please help if you can!!!!
storchak [24]

Answer:Well, I don't know what you got so I can't tell you if it is right.

If it works in both equations, it depends of whether your equations are set up correctly.

Here is how I would do this problem.

Let x = no. of hot dogs,y = number of sodas.

First equation is just about the number of things.

x + y = 15

Second equation is about the cost of things.

1.5 x + .75 y = 18

solve x+y = 15 for y  y = 15-x    substitute into second equation

1.5x + .75(15 - x) = 18    

You should get the correct answer for number of hot dogs if you solve this correctly.  Put your answer in the x + y =15 equation to get y.  Then put both x and y into the cost equation and check your answer.

Hope this helps.

Step-by-step explanation:

5 0
3 years ago
4 - lxl = 1<br> Solve for x
Leokris [45]

Answer: 3 or -3

Step-by-step explanation:

4 - |x| = 1

First subtract 4

-|x| = -3

The | | sign makes it both positive and negative

So -|3| and -|-3| will = -3

4 0
3 years ago
Read 2 more answers
Point E is on line segment DF. Given DE<br> EF.<br> 6 and DF =9, determine the length EF
nlexa [21]

Answer:

<h2>3</h2>

Step-by-step explanation:

IF point E lies on the line segment DF, this means that all the points DEF are collinear and DE+EF = DF.

Given parameter

DE = 6

DF = 9

Required

EF

Substituting the given parameter into the expression above to get the required will be;

DE+EF = DF.

EF = DF-DE

EF = 9-6

EF = 3

Hence the length of EF is equivalent to 3

5 0
3 years ago
a rectangle has an area of 1/6 squar centimeters and a length of 1.5 centimeters what is the width and the perimeter
Alex73 [517]
<span>

(1/6)/ 1.5 = 
(1/6)/(15/10)=
1/6*10/15=10/90=1/9 width</span>
3 0
3 years ago
I alipou
Norma-Jean [14]

Answer:

After the sales tax

$235

before sales tax

210.32

Step-by-step explanation:

3 0
3 years ago
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