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galina1969 [7]
3 years ago
10

For the following system, if you isolated x in the first equation to use the Substitution Method, what expression would you subs

titute into the second equation? -x - 2y = -4 3x + y = 12
a. -2y - 4
b. 2y - 4
c. 2y + 4
d. -2y + 4
Mathematics
1 answer:
tatiyna3 years ago
4 0
-x-2y=-4 \\ 3x+y=12 \\ \\ \hbox{the first equation:} \\ -x-2y=-4 \ \ \ \ \ \ |+2y \\ -x=2y-4 \ \ \ \ \ \ \ \ |\times (-1) \\ x=-2y+4

The expression you would substitute for x is -2y+4. The answer is d.
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How many terms are in the expression <br> 5x^3-8x^2y+4xy-4
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Answer:

<h2>4 terms</h2>

Step-by-step explanation:

Term 1: 5x^3

Term 2: 8x^2y

Term 3: 4xy

Term 4: 4

8 0
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The monomial of the 2nd degree with a coeficcient of 3
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2 years ago
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den301095 [7]

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Mr Wallace sells appliances in receives a weekly salary of $275 plus a 3% Commission on all of his self last week is sells total
Greeley [361]

Answer:

The total earning of Mr Wallace for the week is $310.37  

Step-by-step explanation:

Given as :

The fixed weekly salary of Mr Wallace = $275

And The commission of 3% on his total sell

Last week Mr Wallace total sells amount= $1179

I.e 3% of total sells amount = 3% of $1179 = .03 × $1179 = $35.37

∴The total earning = $275 + $35.37 = $310.37

Hence The total earning of Mr Wallace for the week is $310.37    Answer

8 0
3 years ago
Radioactive Decay Beginning with 16 grams of a radioactive element whose half -life is 30 years ,the mass y(in grams) remaining
Flura [38]

Answer:

the equation should be corrected to fit the data of the problem.  With the corrected equation a mass of 0.5 grams remains after 150 years

Step-by-step explanation:

for the mass y( in grams)

y=23* (1/2)^(t/45), t ≥ 0.

the initial mass is at t=0 , then

y= 23 grams  → should be 16 grams

half-life from the equation = 45 years → should be 30 years

the correct equation should be

y=16*(1/2)^(t/30), t ≥ 0

then after 150 years  → t= 150

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then a mass of 0.5 grams remains after 150 years

5 0
3 years ago
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