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3 years ago

Slope intercept form of y+3= 1/2(x-2)?

Mathematics

1 answer:

WITCHER [35]3 years ago

8 0

The slope intercept form looks like y=mx+b (m-slope, b is the y intercept)

y+3=1/2(x-2)
y=(1/2)x-(2/2)-3

y=(1/2)x-4

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What is equivalent to<br>x2+ 14x = -13

postnew [5]

Answer:

X= 1 AND 13

Step-by-step explanation:

x2+14x = -13

by completing the square method

you add the square of half the coefficient of b to both sides

which is (14/2)^2

x^2+14x+(7^2) = -13 +(7^2)

x^2 + 7^2 = -13 + 49

x^2 + 7^2 = 36

(x + 7)^2 = 36

find the square root of both sides

(x+7) = + or - 6

x=7-6 ; 1

x= 7+6 ; 13

8 0

2 years ago

Find the exact value of cos(sin^-1(-5/13))

son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

$\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}$

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}$

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0

2 years ago

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QveST [7]

Answer: 364 m squared

Step-by-step explanation:

to find the surface area of a rectangular prism, you must do width*length + hight*length + height*width

4 0

2 years ago

Solve square root <br> X^2+12x+43=0

AlekseyPX

$x^2+12x+43=0
$

$x= \dfrac{-b+ \sqrt{b^2-4ac} }{2a}, \dfrac{-b- \sqrt{b^2-4ac} }{2a}$

$x= \dfrac{-12+ \sqrt{12^2-4*43} }{2}$ , $\dfrac{-12- \sqrt{12^2-4*43} }{2}$

$x= \dfrac{-12+2 \sqrt{7i} }{2}, \dfrac{-12-2 \sqrt{7i} }{2}$

$x=-6+ \sqrt{7i},-6- \sqrt{7i}$

5 0

2 years ago

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A regular hexagon is rotated about its center. By which angle could the hexagon be rotated so that it is mapped onto itself?

Sauron [17]

Answer: 60

Step-by-step explanation:

5 0

3 years ago