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2 years ago

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GIVING 20 POINTS NEED HELP

1 answer:

alexandr1967 [171]2 years ago

6 0

Answer:

4 is a BCA def of rectangle

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3. Please show your work to find the mean and standard deviation of the data. Round to the nearest tenth.

Ainat [17]

<span>16.1428571 = mean
3.563 = standard deviation
</span>

7 0

2 years ago

Find the equation of the tangent line to the curve (a lemniscate)

olya-2409 [2.1K]

Answer:

$m=\frac{9}{13}$ and $b=\frac{40}{13}$

Step-by-step explanation:

The equation of curve is

$2(x^2+y^2)^2=25(x^2-y^2)$

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

$2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})$

$4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})$

The point of tangency is (-3,1). It means the slope of tangent is $\frac{dy}{dx}_{(-3,1)}$ .

Substitute x=-3 and y=1 in the above equation.

$4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})$

$40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})$

$-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}$

$80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240$

$130\frac{dy}{dx}=90$

Divide both sides by 130.

$\frac{dy}{dx}=\frac{9}{13}$

If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

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

The slope of tangent line is $\frac{9}{13}$ and it passes through the point (-3,1). So, the equation of tangent is

$y-1=\frac{9}{13}(x-(-3))$

$y-1=\frac{9}{13}(x)+\frac{27}{13}$

Add 1 on both sides.

$y=\frac{9}{13}(x)+\frac{27}{13}+1$

$y=\frac{9}{13}(x)+\frac{40}{13}$

Therefore, $m=\frac{9}{13}$ and $b=\frac{40}{13}$ .

5 0

2 years ago

What is the best approximation for the input value when f(x)=g(x)?

Lostsunrise [7]

Answer:

$x=0$ and $x=1$ .

Step-by-step explanation:

If we have to different functions like the ones attached, one is a parabolic function and the other is a radical function. To know where $f(x)=g(x)$ , we just have to equalize them and find the solution for that equation:

$x^{2}=\sqrt{x} \\(x^{2} )^{2}=(\sqrt{x} )^{2}\\x^{4}=x\\x^{4}-x=0\\x(x^{3}-1)=0\\$

So, applying the zero product property, we have:

$x=0\\x^{3}-1=0\\x^{3}=1\\x=\sqrt[3]{1}=1$

Therefore, these two solutions mean that there are two points where both functions are equal, that is, when $x=0$ and $x=1$ .

So, the input values are $x=0$ and $x=1$ .

8 0

2 years ago

Read 2 more answers

Write the number 11 as a fraction with the denominator of 15.

Elanso [62]

Answer:

11/15

there ya go!!

Step-by-step explanation:

4 0

1 year ago

Read 2 more answers

Is15 a composite or a prime number

Orlov [11]

It is a prime number.

8 0

2 years ago

Read 2 more answers