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3 years ago

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A person has 17 coins, all nickels and dimes, worth $1.25. How many nickels are there?

1 answer:

Leto [7]3 years ago

7 0

<span>9 Nickels, 8 Dimes...here is how i work it out!</span>

d + n = 17
<span>0.1d + 0.05n = 1.25 </span>

<span>d = 17 - n </span>
<span>0.1 (17 - n) + 0.05n = 1.25 </span>
<span>1.7 - 0.1n + 0.05n = 1.25 </span>
<span>1.7 - 0.05n = 1.25 </span>
<span>-0.05n = -0.45 </span>
<span>n = 9 </span>

9 Nickels, 8 Dimes

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Find the midpoint of a segment with the given endpoints.<br> W(-2,5), X(3,-17) =

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The answer is

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Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

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From the question the points are

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The midpoint is

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We have the final answer as

<h3> $( \frac{1}{2} \: , \: - 6)$ </h3>

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3 years ago

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4 years ago

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Let n>=1. <br> Prove by induction for all <br> n>=1 1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1).

alexira [117]

Answer:

Proved

Step-by-step explanation:

Given

$1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)$

$n \geq 1$

Required

Prove by induction

$1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)$

Increment n by 1 on both sides

$1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+1+1)(2(n+1)+1)$

Simplify the right hand side

$1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+2+1)$

$1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+3)$

Group the left hand side

$(1+5+9+...+[4(n-1)+1]+[4n+1])+[4(n+1)+1] = (n+2)(2n+3)$

Recall that

$1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)$ ----[Given]

So; Substitute $(n+1)(2n+1)$ for $1+5+9+...+[4(n-1)+1]+[4n+1]$ on the left hand side

$(n+1)(2n+1)+[4(n+1)+1] = (n+2)(2n+3)$

Open All Brackets

$2n^2 + n + 2n + 1 + 4n + 4 + 1 = 2n^2 + 3n + 4n + 6$

Collect Like Terms

$2n^2 + n + 2n + 4n+ 1 + 4 + 1 = 2n^2 + 3n + 4n + 6$

$2n^2 + 7n+ 6 = 2n^2 + 7n + 6$

<em>Notice that the expression on both sides are equal;</em>

<em>Hence, the given expression has been proven</em>

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3 years ago

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3 years ago