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natka813 [3]
2 years ago
9

Answers choices are in the picture

Mathematics
1 answer:
UkoKoshka [18]2 years ago
6 0

Answer: I believe the answer is b.

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Mary's age is 10 years less than 2 times Joe's age. If the difference between their ages is 30, how old is Mary?
zheka24 [161]

Answer:

its 5

Step-by-step explanation:

cause 2 divided by 30is 15 minus 10 its 5

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You would subtract 90% and 35% and get 55%. You subtract that by 20 and yiu will have 35% of the 90% mixture
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In the figure, the measure of angle 6 is 141°. What is the measure of angle 5?
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Angle 5 is 39 degrees
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Which of the following is equal to the expression (8x)^-2/3 (27x)¯1/3
LiRa [457]

The equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x

<h3>How to evaluate the expression?</h3>

The expression is given as:

(8x)^-2/3 * (27x)^-1/3

Evaluate the exponent 8^-2/3

(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * (27x)^-1/3

Evaluate the exponent (27x)^-1/3

(8x)^-2/3 * (27x)^-1/3 = 1/4(x)^-2/3 * 1/3(x)^-1/3

Multiply 1/4 and 1/3

(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^-2/3 * (x)^-1/3

Evaluate the exponent

(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-2/3 -1/3)

This gives

(8x)^-2/3 * (27x)^-1/3 = 1/12(x)^(-1)

So, we have

(8x)^-2/3 * (27x)^-1/3 = 1/12x

Hence, the equivalent expression of (8x)^-2/3 * (27x)^-1/3 is 1/12x

Read more about equivalent expression at

brainly.com/question/2972832

#SPJ1

3 0
8 months ago
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
2 years ago
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