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3 years ago

14

How do I solve for this?

1 answer:

Alenkasestr [34]3 years ago

8 0

Answer: whats your question?

Step-by-step explanation:

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Derivative of this function<br> <br>Y= -7X³ -8X² +6X +3

zaharov [31]

In general, the derivative of a single term Ax^(n) is A n x^(n-1) .

And the derivative of a sum of many terms is the sum of the derivatives
of the individual terms.

Using these two rules, the derivative (with respect to 'x') of the expression
in the question is . . .

<em> Y' = -21x² - 16x + 6</em>

7 0

2 years ago

Divide. Write in simplest form.<br><br> 2/3 divided by 1/5

Nezavi [6.7K]

Answer:

3 1/3 or 3.3(repeated)

Step-by-step explanation:

2/3 ÷ 1/5

= 2/3 × 5

= 10/3

= 3 1/3 or 3.3(repeated)

I hope this helped!

3 0

2 years ago

Peter and Amanda collect seashells on a beach. They both had 0 seashells when they started. Peter collects 5 seashells every min

ankoles [38]

Answer:

5x+8y

Step-by-step explanation:

Hope Your Thanksgiving Went Well, Here's Some Help Cleaning Up

-TheKoolKid1O1

3 0

3 years ago

Read 2 more answers

Please help me I will really appreciate it!!!

slega [8]

Answer:

<em /> $\displaystyle r=\sqrt[3]{\frac{3V}{4\pi}}$ <em />

<em>The last option matches the correct solution</em>

Step-by-step explanation:

<u>Equation Solving</u>

We have the following equation:

$\displaystyle V=\frac{4}{3}\pi r^3$

And it's required to solve it for r.

Solving this equation needs to perform some operations to have the r isolated from any other variables or constants.

It can be done by conveniently operating on both sides as follows:

Eliminate the denominator by multiplying by 3:

$\displaystyle 3V=4\pi r^3$

Move $4\pi$ to the other side by dividing by $4\pi$ :

$\displaystyle \frac{3V}{4\pi}= r^3$

Swapping sides to have r at the left side:

$\displaystyle r^3=\frac{3V}{4\pi}$

Finally, take the cubic root and leave r alone:

$\displaystyle r=\sqrt[3]{\frac{3V}{4\pi}}$

The last option matches the correct solution

5 0

2 years ago

SOMEONE HELP PLEASE

yan [13]

I have this same question, can anyone answer it?

7 0

3 years ago

Read 2 more answers