Please answer all six questions! : )

Mathematics

1 answer:

stiv31 [10]3 years ago

5 0

$18-(\sqrt[3]{27})^3=18-27=-9$

$\left(\sqrt[3]{-\dfrac18}\right)^3+3\dfrac34=-\dfrac18+\dfrac{15}4=\dfrac{29}8=3\dfrac58$

$5\sqrt[3]{729}-24=5\times9-24=21$

$\dfrac14-2\sqrt[3]{-\dfrac1{216}}=\dfrac14-2\times\left(-\dfrac16\right)=\dfrac7{12}$

$54+\sqrt[3]{-4096}=54-16=38$

$4\sqrt[3]{8000}-6=4\times20-6=74$

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May someone help me with this question and explain?

dybincka [34]

You want to have your x alone so you would get rid of the 3. to do that you must do the opposite of what it's saying which it is multiplying so you would divide 3 and do it on both sides. so 3/3 cancels eachother out and now you have to divide 29 and 3 and that is what x equals

4 0

3 years ago

Read 2 more answers

Can you use the values in Pascal’s triangle to find the powers of 11? Explain.

VladimirAG [237]

Hey!

Hope this helps...

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Yes!

Ultimately (it states) all numbers in existence can be extracted from Pascal's Triangle. Using his binomial formation of: (a + b)^0 = 1 , where y is the number you are using. and (a + b)^x is the equation you use to find the answer of that row... the usage of that row, and the formation of its components is how you find the powers of numbers in a given row...

You would be able to find each row of the Triangle, and by using the association of Multiplication, addition, distribution, and

If you look at the images below you will notice that to find a power of 11 at it's 6th power, you look at the 6th row of Pascal's Triangle... Although not completely relevant, I also added an image of using his Triangle to find the Powers of 2...

Remember, The images below are of only smaller versions of Pascal's Triangle, as his Triangle is of infinite size...