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Anit [1.1K]
3 years ago
13

The product of 5 and an odd number will end in what value? A. 0 B. 1 C. 3 D. 5

Mathematics
2 answers:
Sidana [21]3 years ago
8 0

<u>Answer:</u> The product of 5 and an odd number will end with the value of 5.

<u>Step-by-step explanation:</u>

We are given a number which is '5'

When we multiply the number '5' with a digit, the unit place will have wither '5' or '0'

When the number getting multiplied by '5' is even (0, 2, 4, 6 or 8), then the units place will have 0.

<u>For Example:</u>  Multiplying 62 with 5, we get, (62 × 5) = 310

When the number getting multiplied by '5' is odd (1, 3, 5, 7 or 7), then the units place will have 5.

<u>For Example:</u>  Multiplying 23 with 5, we get, (23 × 5) = 115

Hence, the product of 5 and an odd number will end with the value of 5.

Vera_Pavlovna [14]3 years ago
7 0
I think C 3. hope I helped
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<em>Answer:</em>

<em>r = -</em>\frac{6a}{1-5m^2}<em />

<em>Step-by-step explanation:</em>

<em>Rewrite the equation as </em>\sqrt{x} 6a+r/5r<em>  = m</em>

<em>Remove the radical on the left side of the equation by squaring both sides of the equation.</em>

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<em>Then, you simplify each of the equation. </em>

<em>Rewrite: (</em>\sqrt{\frac{6a+r}{5r} )^{2}<em> as </em>\frac{6a+r}{5r} = m^{2}<em />

<em>Remove any parentheses if needed.</em>

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<em>Multiply each term by r and simplify."</em>

<em>Multiply both sides of the equation by  5.</em>

<em>6a+r= m^2r⋅(5)</em>

<em>Remove parentheses.</em>

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<em>6a+r=5m^2)r</em>

<em>Subtract  5m^2)r  from both sides of the equation.</em>

<em>6a+r-5m^2)r=0</em>

<em>Subtract  6a  from both sides of the equation.</em>

<em>r-5m^2)r=-6a</em>

<em>Factor  r out of  r-5m^2)r  </em>

<em>r(1-5m^2)=-6a</em>

Divide each term by 1-5m^2 and simplify.

r = - \frac{6a}{1-5m^2}

There you go, hope this helps!

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3 years ago
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