3 years ago

(x^b-c)^1÷bc×(x^c-a)^1÷ac×(x^a-b)^1÷ab

Mathematics

1 answer:

stiv31 [10]3 years ago

3 0

Hope this helps (sorry if u cant see that)

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$\frac{14}{35} \times 100\% = 40\%$

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2 years ago

What is the 10th term of the geometric sequence 400, 200, 100...?

Vsevolod [243]

ANSWER

$a_ {10} = \frac{25}{32}$

EXPLANATION

The given geometric sequence is

400, 200, 100...

The first term is

$a_1=400$

The common ratio is

$r = \frac{200}{400} = \frac{1}{2}$

The nth term is

$a_n=a_1( {r}^{n - 1} )$

We substitute the known values to get;

$a_n=400( \frac{1}{2} )^{n - 1}$

$a_ {10} =400( \frac{1}{2} )^{10 - 1}$

$a_ {10} =400( \frac{1}{2} )^{9}$

$a_ {10} = \frac{25}{32}$

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4 years ago

The manager of a store increases the price of a popular product by 5%. Let t be the original price of the product. The new price

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The equivalent expression is 1.05t and then 24 x 0.05 equals 1.2 so 25.2

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3 years ago

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Marat540 [252]

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You invested $11,000 in two accounts paying 3% and 7% annual interest, respectively. If the total interest earned for the year w

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Answer:

I invested $ 10,000 at 7% per year, and $ 1,000 at 3% per year.

Step-by-step explanation:

Given that I invested $ 11,000 in two accounts paying 3% and 7% annual interest, respectively, if the total interest earned for the year was $ 730, to determine how much was invested at each rate, the following calculation must be performed:

5,000 x 0.07 + 6,000 x 0.03 = 530

8,000 x 0.07 + 3,000 x 0.03 = 650

10,000 x 0.07 + 1,000 x 0.03 = 730

Therefore, I invested $ 10,000 at 7% per year, and $ 1,000 at 3% per year.

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3 years ago

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(x^b-c)^1÷bc×(x^c-a)^1÷ac×(x^a-b)^1÷ab​

(x^b-c)^1÷bc×(x^c-a)^1÷ac×(x^a-b)^1÷ab