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gladu [14]
2 years ago
10

Solve for x. Captionless Image x= 6 x = 0 x= 5 x= 10

Mathematics
1 answer:
Pachacha [2.7K]2 years ago
4 0

Answer:

You cannot solve for x on any of these. it will just be the same answer.

Step-by-step explanation:

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Airida [17]
4.415e+8 

Have a wonderful day
8 0
3 years ago
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Substitution method <br><br> x-7y=-9<br> -x+8y=10
max2010maxim [7]

Answer:

x = -2, y=1

Step-by-step explanation:

x-7y=-9---------------------equation 1

-x+8y=10-------------------equation 2

From equation 1, make x the subject of formula

x=7y-9----------------------equation 3

substituting x=7y-9 in equation 2,

-(7y-9)+8y=10

Expanding bracket

-7y+9+8y=10

Collecting like terms

8y-7y=10-9

y=1

substituting y=1 in 3

x=7(1)-9

x=7-9

x=-2

7 0
2 years ago
A line of best fit predicts that when x equals 28 y will equal 27.255 but y actually equals 26 what is the residual in this case
Kobotan [32]

Answer:

The answer is -1.255 for residual value.

Step-by-step explanation:

We are tasked to solve for the residual value given that when x equals 29, y will be equals to 27.255. But, when it is tested, y actual value is 26. The formula in solving residual is shown below:

Residual value = Observed value - predicted value

Residual value =  26 - 27.255

Residual values = -1.255

4 0
2 years ago
What is the volume of papers that the tray can hold?
Vesna [10]
What tray..there is not enough information to help you
6 0
2 years ago
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What are the domain restrictions of the expression g2−7g+10g3−6g2+8g ? Select each correct answer.
Sauron [17]

Answer with Step-by-step explanation:

We are given that an expression

f(g)=\frac{g^2-7g+10}{g^3-6g^2+8g}

We have to find domain restriction of the given function.

f(g)=\frac{(g-5)(g-2)}{g(g-4)(g-2)}

Domain restriction means : It is that value of x when substitute in function then  function will  not defined.

It means it is that values which makes denominator zero.

From given function we can see that

When substitute g=0 then it makes denominator zero.

Hence, the function is not defined at g=0

Substitute g=4

Then, it makes denominator zero.

Hence, function is not defined  at g=4

Substitute g=2

Then,it  makes denominator zero.

Hence, function is not defined  at g=2

Therefore, the function is defined for all values of g except g=0, g=2 and g=4

5 0
3 years ago
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