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Ksenya-84 [330]
2 years ago
14

What is the quotient of 2/5 divided by 3/7A. 15/14B. 14/15C. 35/6D. 6/36​

Mathematics
2 answers:
STALIN [3.7K]2 years ago
8 0

Answer:

= { \tt{ \frac{2}{5}  \div  \frac{3}{7} }} \\  \\  = { \tt{ \frac{2}{5} \times  \frac{7}{3}  }} \\  \\  = { \underline{ \tt{ \:  \:  \frac{14}{15}  \:  \: }}}

<u>Answer</u><u>:</u><u> </u><u>B</u>

Scorpion4ik [409]2 years ago
4 0

2/5 ÷ 3/7

= 2/5 × 7/3

= 2 × 7/5 × 3

= 14/15

Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 3/7 is 7/3) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the next intermediate step, the fraction result cannot be further simplified by canceling.

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Given that the product of 22 integers is equal to 1, prove that their sum is not equal to 0.
Kay [80]

1. If the product of these integers is to be 1, then all of them must be either 1 or -1.

2. Since the product is positive (+1), it must be that there are an *even* number of negative ones (-1), if any.

3. If the sum were 0 it would mean that the number of +1's must equal the number of -1's. So that means there would have to be exactly 22/2=11 of each.

4. But if there were 11 of each, that means the number of -1's would be *odd* and there's no way the product could be +1 (as stated in 2 above).

Hence, the sum is never 0, if the product of 22 integers is equal +1.

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3 years ago
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Factor the expression completely.
evablogger [386]

Answer:

Factoring the expression xy^4+x^4y^4 completely we get \mathbf{xy^4(x+1)(x^2-x+1)}

Step-by-step explanation:

We need to factor the expression xy^4+x^4y^4 completely

We need to find common terms in the expression.

Looking at the expression, we get xy^4 is common in both terms, so we can write:

xy^4+x^4y^4\\=xy^4(1+x^3)

So, taking out the common expression we get: xy^4(1+x^3)

Now, we can factor the term (1+x^3) or we can write (x^3+1) by using formula:

a^3+b^3=(a+b)(a^2-ab+b^2)

So, we get:

xy^4(1+x^3)\\=xy^4(x^3+1)\\Applying\:the\:formula\;of\:a^3+b^3\\=xy^4(x+1)(x^2-x+1)

Therefor factoring the expression xy^4+x^4y^4 completely we get \mathbf{xy^4(x+1)(x^2-x+1)}

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Answer:

22-18-25,

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Using this rule, the possible side lengths out of these choices that could form a triangle are:

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2 years ago
Arianna is preheating her oven before using it to bake. Arianna created the following
fredd [130]

Answer:

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Step-by-step explanation:

7 0
3 years ago
Algebra help?<br><br> 343 = (x/6) ^2/3
Vladimir79 [104]

I can only assume that you meant, "Solve for x:"

Apply the exponent 3/2 to both sides of this equation.  The result will be

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343 = x/6.    

Multiplying both sides by 6 isolates x:

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6*343           = x                         Since 7^3 = 343,  the expression for x

                                                  can be rewritten as

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6*(7^3)          =   x                      which can be further simplified, as follows:


x =  6^(3/2)*7^(9/2), or:

x = 6^(3/2)*7^(8/2)*√7, or

x = 6^(3/2)*7^4*√7

3 0
3 years ago
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