How to solve this equation :

Mathematics

1 answer:

Lady bird [3.3K]2 years ago

6 0

Step-by-step explanation:

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the slope of the line below is -3. Use the coordinates of the labeled point to find a point-slope equation of the line

zhenek [66]

The answer to the question is a

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4 years ago

What is the value, to the nearest tenth, of 15 x 3 - 12 x 2 when x = 2.5

Alekssandra [29.7K]

Answer:

159.4

Step-by-step explanation:

I'm assuming that it's supposed to be 15x^3-12x^2 because when you copy and paste the "^" gets cut out

15*2.5^3 - 12*2.5^2 = 234.375 - 75 = 159.375

rounded is 159.4

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3 years ago

AP Calculus Help Please!

ivanzaharov [21]

4. By the fundamental theorem of calculus,

$\displaystyle\frac{\mathrm dg(x)}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\int_{-2}^xf(t)\,\mathrm dt=f(x)$

$g(x)$ is increasing on those intervals where $g'(x)=\dfrac{\mathrm dg(x)}{\mathrm dx}>0$ . So you have

$g'(x)=f(x)=\begin{cases}3&\text{for }-3\le x$

which is clearly positive for $x\in[-3,0)$ , and in the second interval you have

$-x+3>0\implies x$

Together, this means $g'(x)>0$ for all $x\in[-3,3)$ .

5. When $0\le x\le6$ , $f(x)$ reduces to $-x+3$ , so you have

$g(x)=\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt$
$g(x)=6+\left(3t-\dfrac12t^2\right)\bigg|_{t=0}^{t=x}$
$g(x)=6+3x-\dfrac12x^2$