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3 years ago

What is 8x+4y=24 in slope intercept form.

Mathematics

2 answers:

Vikentia [17]3 years ago

7 0

Y=-2x+12 is slope intercept form. -2 is the slope and 12 is the y intercept

sergeinik [125]3 years ago

5 0

Answer:

Slope: 2y- int: −6

Explanation:

We can easily find the slope and

y -intercept of this line by converting it to slope intercept form

y=mx + b, with slope m and a

y- intercept of b.

Let's subtract

from both sides to get

−4y=−8x+24

Next, we divide all terms by

−4 to get y=2x−6

Now that our equation is in this form, we see that our slope is

2, and our y -intercept is −6.

-Hope this helps you

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Which is smaller,two-thirds of 27 or two-fifths of 20?

shtirl [24]

" of " means multiply

2/3(27) = 54/3 = 18
2/5(20) = 40/5 = 8

so 2/5 of 20 is smaller

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3 years ago

Math help please TY :3 HWAJDSK 10 POINTS REWARDING !!

valkas [14]

I will do Point A carefully, The others I will indicate. Start with the Given Point A. Then do the translations

A(-1,2) Original Point

Reflection: about x axis:x stays the same; y becomes -y:Result(-1,-2)

T<-3,4>: x goes three left, y goes 4 up (-1 - 3, -2 + 4): Result(-4,2)

R90 CCW: Point (x,y) becomes (-y , x ) So (-4,2) becomes(-2, - 4): Result (-2, - 4)

B(4,2) Original Point

Reflection: (4, - 2)
T< (-3,4): (4-3,-2 + 4): (1 , 2)
R90 CCW: (-y,x) = (-2 , 1)

C(4, -5) Original Point

Reflection (4,5)
T<-3,4): (4 - 3, 5 + 4): (1,9)
R90, CCW (-9 , 1)

D(-1 , -5) Original Point

Reflection (-1,5)
T(<-3,4): (-1 - 3, 5 + 4): (-4,9)
R90, CCW ( - 9, - 4)

Note: CCW means Counter Clockwise

The graph on the left is the same one you have been given.

The graph on the right is the same figure after all the transformations

7 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago