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3 years ago

How would the expression x^3+64 be rewritten using sum off cubes

2 answers:

8 0

$\bf \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\
a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\\\\
-------------------------------\\\\
\boxed{64=4^3}\qquad x^3+64\implies x^3+4^3\implies (x+4)(x^2-x4+4^2)
\\\\\\
(x+4)(x^2-4x+16)$

gogolik [260]3 years ago

3 0

Answer:

$(x+4)(x^2-4x+4^2)$

Step-by-step explanation:

We have been given an expression $x^3+64$ and we are asked to rewrite our expression using sum of cubes.

Sum of cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$ .

We can rewrite 64 as: $64=(4*4*4)=4^3$

This means that a = x and b = 4, Upon substituting these values in sum of cubes formula we will get,

$x^3+64=(x+4)(x^2-4x+4^2)$

Therefore, after rewriting our given expression as sum of cubes we will get: $(x+4)(x^2-4x+4^2)$ .

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The sum of two polynomials is 8d^5-3c^3d^2+5c^2d^3-4CD^4+9. If one added is 2d^5-c^3d^4+8CD^4+1, what is the other added

insens350 [35]

Answer:

6d^5 - 3c^3d^2 + 5c^2d^3 + c^3d^4 - 12cd^4 + 8

Step-by-step explanation:

We need to subtract the given polynomial from the sum:-

8d^5 - 3c^3d^2 + 5c^2d^3 - 4cd^4 + 9 - (2d^5 - c^3d^4 + 8cd^4 +1 )

We need to distribute the negative over the parentheses:-

= 8d^5 - 3c^3d^2 + 5c^2d^3 - 4cd^4 + 9 - 2d^5 + c^3d^4 - 8cd^4 -1

Bringing like terms together:

= 8d^5 - 2d^5 - 3c^3d^2 + 5c^2d^3 + c^3d^4 - 4cd^4 - 8cd^4 + 9

- 1

Simplifying like terms

= 6d^5 - 3c^3d^2 + 5c^2d^3 + c^3d^4 - 12cd^4 + 8

8 0

3 years ago

Help me out please i need this

mart [117]

Answer:

Just restart your computer

8 0

3 years ago

**Spam answers will not be tolerated**

Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

3 years ago

Glenda and Penny are doing

Volgvan

Answer:

It depends on how many inches they are from the finish line.

Step-by-step explanation:

However, they will be at the same distance at 90 inches when Glenda hops 5 times. 18 x 5 = 90. Penny will also make it to 90 inches when she hops 6 times. 15 x 6 = 90. Then, you will subtract 90 inches from how long the race is in inches.

Hope this helps!

4 0

3 years ago

The area of the triangle below is 2/5 square foot. What is the length in feet of the base of the triangle?

solmaris [256]

Answer:

C) 2/3

Step-by-step explanation:

8 0

3 years ago