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MA_775_DIABLO [31]

3 years ago

12

Solve for x.x = ___273330

2 answers:

LenKa [72]3 years ago

8 0

The answer is 27.

The answer needs to equal 90°(right angle)
3(27)+9
81+9
=90

Hope this helps! :)

kirill [66]3 years ago

4 0

The answer is Answer 30 i think

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Need help ASAPPPPPPP

patriot [66]

Answer:

8 is the correct answer because coefficient lies before Exponent. Exponent is variable

So 8 is the coefficient.

5 0

2 years ago

If AB = x+ 4, BC = 2x -10 and AC

lawyer [7]

Answer:

15

Step-by-step explanation:

By the diagram, you can see the sum of segment AB and segment BC is segment AC. Adding the given expressions for AB and B is $3x-6$ . Simplifying the equation $3x-6=2x+1$ , gives $x=7$ . Substituting $x=7$ in the equation for segment AC gives 15.

8 0

2 years ago

PLEASE HELP ONE QUESTION!!!!!! PLEASEEEEE

In-s [12.5K]

VW=CD since they are congruent figure, so CD=6

7 0

3 years ago

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Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago

Which logarithmic equation is equivalent to the exponential equation below?

jek_recluse [69]

The answer is B.
ln both sides to clear the 'e' and bring 'a' out of the power.

7 0

3 years ago

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