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Stels [109]
3 years ago
5

Find the product. (do not use spaces in your answer.) d ^j · d ^k

Mathematics
2 answers:
olganol [36]3 years ago
6 0

Answer: The product would be d^{j+k}

Step-by-step explanation:

Since we have given that

d^j.d^k

We need to find the product.

As we know that

a^m\times a^n=a^{m+n}

So, we will use the above exponential law and it becomes,

d^j.d^k=d^{j+k}

Hence, the product would be d^{j+k}

bezimeni [28]3 years ago
3 0
d^j\cdot d^k=d^{j+k}
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A collection of 30 coins worth $5.50 consists of nickels, dimes and quarters. There are twice as many dimes as nickels. How many
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<span>Let N equal the number of nickels.
Let D equal the number of dimes.
Let Q equal the number of quarters.

We know that .05N + .10D + .25Q = 5.50.
We also know that N + D + Q = 30.
We also know that D = 2N.

Therefore, Q = 30 - (N + D) = 30 - (N + 2N) = 30 - 3N

So...

5.50 = .05N + .10(2N) + .25(30 - 3N)
= .05N + .20N + 7.50 - .75N

Therefore (moving all N's to the left of the equals sing, and all constants to the right of the equals sign)...

.5N = 2.

Therefore, N = 4. There are four nickels, eight dimes, and 18 quarters.

Checking our work...
Four nickels is 0.20.
Eight dimes is 0.80.
18 quarters is 4.50.

That sums to 5.50, and is 30 coins.</span>
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