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Stolb23 [73]
3 years ago
7

If 8 is added to twice a number and this sum is multiplied by 5​, the result is the same as if the number is multiplied by negat

ive −28 and 2 is added to the product. what is the​ number?
Mathematics
1 answer:
Naily [24]3 years ago
3 0
(8+2x) -> If 8 is added to twice a number (x)
5*(8+2x) -> and this sum is multiplied by 5
5*(8+2x)=x*(-28)+2   -><span>the result is the same as if the number is multiplied by negative −28 and 2 is added to the product.

Simplifying the equations on both sides would give
40+10x=-28x+2 transpose -28 to the left side and 40 to the right side
28x+10x=2-40
38x=-38
x=-1</span>
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mamaluj [8]

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One solution

Step-by-step explanation:

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The cost of 6one by two metere
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Rs 17 3/4

Step-by-step explanation:

Cost of 6 1/2 m = 115 3/8

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5 0
3 years ago
How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?
MrRissso [65]
\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3}  \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ =  \frac{(3x)^{3 \times  \frac{5}{3} }}{(2y^3)^{3 \times  \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}

Now, If the exponent was negative like you asked....

\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}

5 0
3 years ago
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krek1111 [17]

Answer:

20

Step-by-step explanation:

4/7 is the same as 4 divided by 7. When you do that you get a really long decimal, I made sure I was right by using a calculator and multiplying that by 35 to get 20.

Hope that helps and have a great day!

7 0
3 years ago
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