3 years ago

Please answer this question!!

Mathematics

2 answers:

HACTEHA [7]3 years ago

7 0

Yes it is I’m not sure tho the other person will help

Pachacha [2.7K]3 years ago

7 0

That is NOT correct. This is because:

(1/3)^2 = 1/9
(1/2)^3 = 1/8

So, that equation up there is not true and his reasoning is wrong.

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One positive number exceeds three times another positive number by 5. The product of the two numbers is 68. Find the number

kati45 [8]

Answer:12

Step-by-step explanation:

Let the numbers be a and b

a-3b=5 ........equation (1)

There product is :

ab=68 ..........equation (2)

From equation (1)

a-3b=5

Make a the subject of the formula

a=5+3b. ........equation (3)

Substitute equation (3) into equation (2)

(5+3b)b=68

3(b^2)+3b-68=0

The product is -204b^2(17 and -12)

(b+17)(b-12)=0

b=-17 or b=12

So the number is 12

6 0

3 years ago

Simplify the expression below, I really just need the steps I have the answer (also can someone tell me how to edit points on th

Sloan [31]

Answer: $3x^2y\sqrt[3]{y}\\\\$

Work Shown:

$\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\$

Explanation:

As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.

4 0

3 years ago

Fill in the missing exponent to make the equation true. 59 - 50-53 No links .

andrezito [222]

Answer:

the answer is 6.

Step-by-step explanation:

$5^{9}/5^{6}=5^{3}$