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2 years ago

Lol i have 69 points GOOBA STOOPID

Mathematics

2 answers:

krok68 [10]2 years ago

6 0

Answer:

okay thank you for the points

amm18122 years ago

3 0

Answer:

:D HUh

Step-by-step explanation:

THANKS FOR THE POINTS !!!!!

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A triangle has one side length of 12 inches and another of 8 inches. Describe all possible lengths of the third side. Show and e

katrin [286]

Answer:

The third side must be smaller than 20 inches and greater than 4 inches

$4$

Step-by-step explanation:

Let a, b and c be the lengths of triangle's sides. Then

$a+b>c\\ \\a+c>b\\ \\b+c>a$

Use this rule in your case. So, if a=12 and b=8, then

$12+8>c\\ \\12+c>8\\ \\8+c>12$

Hence, you get

$c-4\\ \\c>4$

From these inequalities, you can state that

$4$

So, c must be smaller than 20 inches and greater than 4 inches.

5 0

3 years ago

Read 2 more answers

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

Helppppp please!! :/

Elan Coil [88]

It's 3.00 boom son gghjjj

6 0

3 years ago

I need help is it A B C or D

Evgesh-ka [11]

Answer is A, different slopes
enjoy

6 0

2 years ago

Read 2 more answers

Is (5x+4) a factor of (-21x^2-15-40x+5x^3

Tamiku [17]

Answer:

No. It is not.

Step-by-step explanation:

Dividing 5x³+21x²-40x-15 by 5x+4 gives x²+5x-4 with a remainder. Which means, 5x+4 is not a factor of 5x³+21x²-40x-15

6 0

3 years ago