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Ghella [55]
3 years ago
7

Can some one pls help

Mathematics
1 answer:
siniylev [52]3 years ago
6 0

Answer:

m=9

Step-by-step explanation:

(y2-y1)/(x2-x1)=m

In your scenario:

(7-(-2))/(-6-(-7))=m

m=9

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What is the result of an increase by 2/3
velikii [3]

Answer:

y=1.6 repeatin

Step-by-step explanation:

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The diameter of a circle is 8 inches. What is the area? 8 pi iThe diameter of a circle is 8 inches. What is the area? 8 pi inche
Aleksandr [31]

Answer:

16pi inches squared is the answer

Step-by-step explanation:

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3 years ago
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Given f(x) = (lnx)^3 find the line tangent to f at x = 3
kirill [66]
Explanation

We must the tangent line at x = 3 of the function:

f(x)=(\ln x)^3.

The tangent line is given by:

y=m*(x-h)+k.

Where:

• m is the slope of the tangent line of f(x) at x = h,

,

• k = f(h) is the value of the function at x = h.

In this case, we have h = 3.

1) First, we compute the derivative of f(x):

f^{\prime}(x)=\frac{d}{dx}((\ln x)^3)=3*(\ln x)^2*\frac{d}{dx}(\ln x)=3*(\ln x)^2*\frac{1}{x}=\frac{3(\ln x)^2}{x}.

2) By evaluating the result of f'(x) at x = h = 3, we get:

m=f^{\prime}(3)=\frac{3}{3}*(\ln3)^2=(\ln3)^2.

3) The value of k is:

k=f(3)=(\ln3)^3

4) Replacing the values of m, h and k in the general equation of the tangent line, we get:

y=(\ln3)^2*(x-3)+(\ln3)^3.

Plotting the function f(x) and the tangent line we verify that our result is correct:

Answer

The equation of the tangent line to f(x) and x = 3 is:

y=(\ln3)^2*(x-3)+(\ln3)^3

6 0
1 year ago
Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4
Alik [6]

Answer:

\alpha=3

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

(x-h)^2+(y-k)^2=r^2

We are given the equation of a circle as

3x^2+3y^2-6\alpha x+12y-3\alpha=0

Note we have corrected it by adding the square to the y. Simplify by 3

x^2+y^2-2\alpha x+4y-\alpha=0

Complete squares and rearrange:

x^2-2\alpha x+y^2+4y=\alpha

x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4

(x-\alpha)^2+(y+2)^2=r^2

We can see that, if r=4, then

\alpha+\alpha^2+4=16

Or, equivalently

\alpha^2+\alpha-12=0

There are two solutions for \alpha:

\alpha=-4,\ \alpha=3

Keeping the positive solution, as required:

\boxed{\alpha=3}

8 0
3 years ago
HELP PLS!!!
kotykmax [81]

Answer:

B. \frac{60 hours}{1} \times \frac{1 day}{24 hours}

Step-by-step explanation:

Converting 60 hours to days:

24 hours = 1 day

60 hours = x day

Cross multiply

24 hours \times x = 60 hours \times 1 day

Divide both sides by 24 hours

x = \frac{60 hours \times 1 day}{24 hours}

Thus,

\frac{60 hours \times 1 day}{24 hours} = \frac{60 hours}{1} \times \frac{1 day}{24 hours}

The expression that shows how to convert 60 hours to days is:

\frac{60 hours}{1} \times \frac{1 day}{24 hours}

5 0
3 years ago
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