How do I find the derivative of the problem: f(x) = x^2 + x at x= -1

In a zoo there are 21 Lions. 1/3 of them are male. How many females are in the zoo?

Jet001 [13]

Answer: 14 female
Explanation:
21/3= 7 male
21-7=14 female
hope it helps!

5 0

2 years ago

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What is the formula for a counterclockwise rotation of 180°?!?!

Sloan [31]

Answer: A (x,y) is your original. Your answer after would be A' (-x,-y) you would change the sign. If its negative it becomes positive, and if it'd positive it becomes negative.

7 0

2 years ago

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NEED HELP ASAP GIVING BRANILEST

finlep [7]

Answer:

50?

Step-by-step explanation:

I said 50 because there were 60 na butterflies so there was probably 50.

I am sorry if this wrong tho

7 0

2 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

Hellllllllllllllllllllllllllp

murzikaleks [220]

Answer:

$\begin{cases}y=-5x+1\\y=5x-4 \end{cases}$

Step-by-step explanation:

Slope-intercept form of a linear equation:

$\boxed{y=mx+b}$

where:

m is the slope.
b is the y-intercept (where the line crosses the y-axis).

Slope formula

$\boxed{\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}}$

Equation 1

Define two points on the line:

$\textsf{Let }(x_1,y_1)=(-1, 6)$

$\textsf{Let }(x_2,y_2)=(0, 1)$

Substitute the defined points into the slope formula:

$\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1-6}{0-(-1)}=-5$

From inspection of the graph, the line crosses the y-axis at y = 1 and so the y-intercept is 1.

Substitute the found slope and y-intercept into the slope-intercept formula to create an equation for the line:

$y=-5x+1$

Equation 2

Define two points on the line:

$\textsf{Let }(x_1,y_1)=(1, 1)$

$\textsf{Let }(x_2,y_2)=(0, -4)$

Substitute the defined points into the slope formula:

$\implies \textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-4-1}{0-1}=5$

From inspection of the graph, the line crosses the y-axis at y = -4 and so the y-intercept is -4.

Substitute the found slope and y-intercept into the slope-intercept formula to create an equation for the line:

$y=5x-4$

Conclusion

Therefore, the system of linear equations shown by the graph is:

$\begin{cases}y=-5x+1\\y=5x-4 \end{cases}$

Learn more about systems of linear equations here:

brainly.com/question/28164947

brainly.com/question/28093918

5 0

1 year ago

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