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3 years ago

ME NEEEEEEEED HE;LP NOOOOOOOOOW pls and swer even if u can only answer 1 it wud be appreciated, ty

Mathematics

1 answer:

Kay [80]3 years ago

8 0

Answer:

x ≤ -3/2 or x≥1/3

Step-by-step explanation:

-6x^2-7x+3≤0

-(3x-1)(2x+3)≤0

*-1 ≤ *-1

x ≤ -3/2 or x≥1/3

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Is y=3x-5 and 6x=2y+10 a solution

iVinArrow [24]

Y=3x-5
y-3x=-5
Times 2 from both side
y(2)-3x(2)=-5(2)
2y-6x=-10
Times negative from both side
(-)2y-6x(-)=(-)-10
-2y+6x=10
or
6x-2y=10

6x=2y+10
6x-2y=10
Furthermore, we see that both equation have the same 6x-2y=10 which means that both of them have infinity solutions, not a solution. Hope it help!

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3 years ago

Use the x-intercept method to find all real solutions of the equation. -9x^3-72x^2+36=3x^3+x^2-3x+8

larisa86 [58]

-9x^3-72x^2+36=3x^3+x^2-3x+8 Add 9x^3 to both sides.

-72x^2 + 36 = 3x^3 + 9x^3 + x^2 - 3x + 8 Add 72x^2 to both sides

36 = 12x^3 + 73x^2 - 3x + 8 Subtract 36 from both sides.

0 = 12x^3 + 73x^2 - 3x - 28

It does factor, but it is not very nice.

(x + 6.06)(x - 6.09)(x + 0.632)

If there is any kind of error please report it in a note below.

6 0

3 years ago

What’s the answer? Please Please Help me

OverLord2011 [107]

The answer is 155 . 234-79=155 !

5 0

3 years ago

Read 2 more answers

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

3(х + 3) 3х + 3 please help

shtirl [24]

I don’t really know but try a calculator

4 0

3 years ago