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3 years ago

I need help big time. Will reward BRAINLIEST to BEST answer!!!

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

The slope of the line in simplified form is -18.

Kisachek [45]3 years ago

4 0

The find the equation of a line in any form, you will have to find out the slope first:
m=(y2-y1)/(x2-x1)=(48-12)/(32-34)=-18
the simplified form is the slope-intercept form, so next we need to find the y intercept,
y=mx+b =>y=-18x+b
Use any of the two given points to find out b: 12=-18(34)+b =>b=624
so the equation is: y=-18x +624
Please double check the calculation by yourself

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What is the approximate area of the circle shown below?

Lynna [10]

Answer:

A. 2827 square cm

Step-by-step explanation:

Diameter of circle = 60 cm

Therefore, radius = 60/2 = 30 cm

$Area \: of \: circle \\ = \pi {r}^{2} \\ = 3.14159265\times {30}^{2} \\ = 3.14159265 \times 900 \\ = 2,827.43339 \: {cm}^{2} \\ = 2827 \: {cm}^{2}$

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3 years ago

Factor x^2 + 10x -18

vovangra [49]

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4 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

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4 years ago

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3 years ago

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umka21 [38]

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