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3 years ago

3x^2-y^3-y^3-z if x=3, y=-2, and z=-5

Mathematics

2 answers:

Serggg [28]3 years ago

5 0

$3(3)^2-(-2)^3-(-2)^3-(-5) = 48$

48 is the answer.

WITCHER [35]3 years ago

4 0

48 is the answer!!!!!!!

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URGENT please help Find part B

lyudmila [28]

Answer:

b = 45 / 4 = 11 1 /4.

Step-by-step explanation:

We are dealing with a (30,60,90)-triangle, therefore z = 15 sqrt(3) / 2. So as well b = (15 sqrt(3) / 2) sqrt(3) / 2 = 45 / 4.

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Bananas cost $2 a bunch and apples cost $0.75 each. If “b” represents the number of bunches of bananas and “a” represents the nu

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Answer: 2b + 0.75a

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Calc help pretty please

Gre4nikov [31]

Answer:

A. 77.4

Step-by-step explanation:

Linear approximation formula

$L(x)=f(a)+f'(a)(x-a)$

Given function:

$f(x)=(x+1)^2$

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times \dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

$\boxed{\begin{minipage}{4 cm}\underline{Differentiating $ax$}\\\\If $y=ax$, then $\dfrac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}$

Use the chain rule to differentiate the function.

$\textsf{Let }\:y=u^2\:\textsf{ where }u=(x+1)$

Differentiate the two parts separately:

$y=u^2 \implies \dfrac{\text{d}y}{\text{d}u}=2u$

$u=x+1 \implies \dfrac{\text{d}u}{\text{d}x}=1$

Put everything back into the chain rule formula:

$\begin{aligned} \implies \dfrac{\text{d}y}{\text{d}x} & =2u \times 1\\ & = 2u \\ & = 2(x+1)\\ & = 2x+2 \end{aligned}$

$\textsf{Therefore, }\:f'(x)=2x+2.$ .

The linear approximation at a = 8 is:

$\begin{aligned}L(x) & =f(a)+f'(a)(x-a)\\\\\implies L(x) & = f(8)+f'(8)(x-8)\\& = (8+1)^2+(2(8)+2)(x-8)\\& = 81+18(x-8)\\& = 18x-63\end{aligned}$

Finally, substitute x = 7.8 into the linear approximation equation:

$\begin{aligned}\implies L(7.8) & =18(7.8)-63\\& = 140.4-63\\& = 77.4\end{aligned}$

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