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Natali [406]
3 years ago
15

For the following system, if you isolated x in the second equation to use the substitution method, what expression would you sub

stitute into the first equation?
3x + y = 8
−x − 2y = −10

A.) −2y + 10
B.) 2y + 10
C.) 2y − 10
D.) −2y − 10
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

A.) −2y + 10

Step-by-step explanation:

−x − 2y = −10

Add 2y to both sides.

-x = 2y - 10

Multiply both sides by -1.

x = -2y + 10

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How can I solve this? and can you please tell me how step by step.
Ludmilka [50]

Answer:27/343 or 0.0787172

Step-by-step explanation:

First you need to place the exponent on both the numerator and the demonimator to make 3 to the power of 3 over 7 to the power of 3

Next you evaluate the numerator so 3 to the power of 3 is 27

and 7 to the power of 3 is 343 so your answer in fraction form is 27/343

you can also change into a decimal if you divide the fraction if you need to

Hope this helps

8 0
3 years ago
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Help me on this pleasseeee
EastWind [94]

Answer:

1) E and F

Step-by-step explanation:

So width is x

legth Is 3x+3

Perimeter is

3x+3+3x+3+x+x

Now simplify

8x+6

Now go through each choice that equals 8x+6

Only e and f

8 0
2 years ago
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Find the product -2X²×4X²y²​
nydimaria [60]

Answer:

-8X^4y^2

Step-by-step explanation:

-2X² × 4X²y²​ = -8X^4y^2

edit to add:

first: multiply the coefficients together:

-2 x 4 = -8

second: add the exponents of X together (when you multiply exponents, you just add them together):

² & ² = ^4

and you have the answer:

-8X^4y^2

6 0
2 years ago
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If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes
Keith_Richards [23]
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
4 0
3 years ago
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kirza4 [7]
Radius is 5 cm. is the answer

8 0
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