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2 years ago

What is the answer to both and how am i suposed to do it

Mathematics

1 answer:

Feliz [49]2 years ago

6 0

First you could switch the order of the numbers were the two fractions are that would now be 1/4 + 4/3(-12)
when you add 1/4 + 4/3 you would get 1 7/12. then you would do 1 7/12 x -12. Multiplying is easy just go straight across. but since its whole and mixed number you would put a one under the -12.

1 7/12 x -12 is -19

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Find the derivative of g(y)=(y-4)*(2y+y^2)

sasho [114]

Answer:

$g'(y)=3y^2-4y-8$

Step-by-step explanation:

start by foiling out the given function

$g(y)=(y-4)(2y+y^2)\\=2y^2+y^3-8y-4y^2\\=y^3-2y^2-8y$

next, use the power rule to find the derivative

power rule: To use the power rule, multiply the variable's exponent n, by its coefficient a, then subtract 1 from the exponent. If there's no coefficient (the coefficient is 1), then the exponent will become the new coefficient.

$g'(y)=3y^2-4y-8$

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1 year ago

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riadik2000 [5.3K]

(
x
+
6
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(
x
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8
)
Is the awnser for this problem

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2 years ago

What is the range of the function y = x2?

den301095 [7]

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2 years ago

Please Help<br> Question; What is the value of x?

Maksim231197 [3]

Answer:

x+5/30 = 16/20

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Step-by-step explanation:

x+5/30 = 16/20

20x+100=480

20x=380

x=19

6 0

2 years ago

25 POINTS AND BRAINLIEST PLEASE HELP ASAP

dexar [7]

M is a midpoint of BC so:

$M=\left(\dfrac{\boxed{2}\boxed{a}+a}{\boxed{2}},\dfrac{\boxed{0}+b}{2}\right)=\left(\dfrac{\boxed{3}\boxed{a}}{\boxed{2}},\dfrac{\boxed{b}}{\boxed{2}}\right)$

Length of MA:

$MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=
\sqrt{\left(\dfrac{\boxed{3}a}{\boxed{2}}\right)^2+\left(\dfrac{b}{2}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

Length of NB:

$NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\
\sqrt{\left(\dfrac{-3a}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

5 0

3 years ago

Read 2 more answers