NEED HELP IMMEDIATELY

Mathematics

2 answers:

lesya [120]3 years ago

5 0

Answer:

(13, −1)

Step-by-step explanation:

Both figures will have the same transformation applied.

We can see from JKLM and J'K'L'M' that the transformation is a translation 7 units right and 5 units up.

Point T has coordinates (6, -6). Applying this translation, we would have T' at (6+7, -6+5) = (13, -1).

Karo-lina-s [1.5K]3 years ago

4 0

I'm sorry, but this doesn't make sense. Point T is nowhere near any of the given coordinates. I could be really bad at graphs, for all I know, but I'm just telling you what I know.

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Vera_Pavlovna [14]

Answer:

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Step-by-step explanation:

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2 years ago

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Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$