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r-ruslan [8.4K]
3 years ago
8

Can someone please help

Mathematics
2 answers:
Nata [24]3 years ago
5 0
F(x)= x^2-3x+4
f(-2)= (-2)^2 - 3(-2) + 4
= 4+6+4
=14
klemol [59]3 years ago
4 0
You just plug in the -2 for x to get the equation -2^2–3•-2+4= 6
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Weight of copper: 62,000 lbs. (31 tons). Weight of framework: 250,000 lbs. (125 tons). Weight of concrete foundation: 54,000,000 lbs. Thickness of copper sheeting: 3/32 of an inch, the thickness of two pennies placed together.  the answer in 125

Step-by-step explanation:

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2 years ago
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For 4 + N years = (1 + 2/100)^(4+N)

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4 0
4 years ago
A line passes through the point (-2, 7) and has a slope of -5.
Ludmilka [50]

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Step-by-step explanation:

3 0
3 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
Evaluate 3b2 when b= 4 and when b= - 4.<br> When b=4, 3b2
nydimaria [60]

Answer:

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Step-by-step explanation:

3 times 4 times 2

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Hope this helps you

4 0
3 years ago
Read 2 more answers
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