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2 years ago

6

Solve linear equations 1/2x+y=-6 and y=3/5x+5

1 answer:

ser-zykov [4K]2 years ago

5 0

1/2x+y=-6 y=3/5x+5

substitute
1/2x+3/5x+5=-6

11/10x+5=-6

11/10x=-11
11x=-110
x=-10

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Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8.

Gala2k [10]

Vertex is directly in middle of directix and focus

distance from 8 to -8 is 16
16/2=8
so 8 below focus (since 8>-8) is the point (0,0
vertex is (0,0)
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also it goes up down so
4p(y-k)=(x-h)^2
(h,k) is veretx
we got that (h,k) is (0,0)
and p is distance from vertex to focus which is 8
so
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3 years ago

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vova2212 [387]

Answer:

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2 years ago

What is the scientific notation for 0.0138

xxMikexx [17]

Answer:

1.38 * 10^ -2

Step-by-step explanation:

.0138

We need 1 number to the left of the decimal

1.38 * 10^ something

We moved it 2 places to the right, that means our exponent is -2. If we had moved it to the left, our exponent would have been positive

1.38 * 10^ -2

8 0

3 years ago

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

AVprozaik [17]

Step-by-step explanation:

Y is directly proportional to X" can be rewritten as a mathematical expression of the form Y = KX

"K" is the constant of proportionality and must be determined from the initial information.

If Y = 2/3 when X=1/12means 2/3=1/12

K = 2/3÷1/12 = 8

Now that we know the value of K, we can use it to calculate the value of Y when X=1/2

Y = (8)(1/2) = 8/2=4

Y=4

8 0

2 years ago