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4 years ago

Plz help

1 answer:

3 0

I'm just going to take a dab at this. I haven't done calculus in years.

So you found the equation of the tangent line for your equation at the point (1,2) which is:
y = 5 - 3x

Which is correct and all, but the question seems to be asking for "another tangent", so we are looking for another tangent line that is parallel to:
y = 5 - 3x

So we know that the equation of this line must also have a slope of "-3".

So if it must have a slope of "-3", then it must follow that:

-3 = -3 + 24x - 24x²

I got the above from setting our derivative equation equal to -3.

So if we solve the above equation we get:

24x - 24x² = 0
24x(1 - x) = 0

So "x" must be equal to 0 or 1. (These are the two points that have a slope of -3)
And since we already know about the point x = 1, we are interested in the point x = 0. So from our original equation the point x = 0 has a y-value of 1, therefore we have the point (0,1). Finally, we just need to find the equation of the line that has a slope of -3 and contains the point (0,1).

I'm sure you can do this part on your own. The equation of that line is:

y = -3x +1
or
y = 1 - 3x (as you pointed out)

Whew, that was a nice refresher. :)

I hope that was easy to follow!

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Question -

Sauron [17]

$\rule{200}4$

Answer : The required ratio is (14m-6):(8m+23) .

$\rule{200}4$

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\$

As , we know that the sum of n terms of an AP is given by ,

$\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\$

Assume that ,

First term of 1st AP = a
First term of 2nd AP = a'
Common difference of 1st AP = d
Common difference of 2nd AP = d'

Using this we have ,

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\$

$\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\$

Now also we know that the nth term of an AP is given by ,

$\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\$

Therefore,

$\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\$

$\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\$

From equation (i) and (iii) ,

$\longrightarrow\sf\small n-1 = 2m-2\\$

$\longrightarrow\sf\small n = 2m -2+1 \\$

$\longrightarrow\sf\small n = 2m -1 \\$

Substitute this value in equation (i) ,

$\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\$

Simplify,

$\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\$

$\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

$\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

From equation (ii) ,

$\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\$

$\rule{200}4$

8 0

3 years ago

Which equation represents the function shown in the input-output table?

stepladder [879]

Answer:

Equation B

Step-by-step explanation:

Equation B because:

it is 7x + 3

so for the first one, it's 7 + 3

the second one is 14 + 3

the third one is 21 + 3

the fourth one is 28 + 3

Please Mark Brainliest

4 0

3 years ago

What is the domain and range of the set of order pairs? (7,3),(-2,-2),(4,1),(-9,0),(0,7)

erastovalidia [21]

Answer:

Domain {7,-2,4,-9,0} Range {3,-2,1,0,7}

Step-by-step explanation:

Just remember that domain is the x values and the range is the y values.

5 0

3 years ago

Explain how to isolate the variable in the equation -2/3+7=15

Tasya [4]

Answer:

n= -12

Step-by-step explanation:

- 2n/3 + 7=15 Subtract 7 from both sides and move all terms containing

-7 = -7 n to the right

- 2n/3 = 8 multiply both sides of equation by - 3/2

-3/2 x (-2n/3) = -3/2 x 8

n= -3/2 x 8 cancel the common factor of 3

n= -12

8 0

3 years ago

Triangle ABC is shown below. Using a compass and straightedge, construct the dilation of ABC centered at B with a scale factor o

qaws [65]

Triangle ABC would remain the same because a dilation generally preserves the measure of an angle.

<h3>What is scale factor?</h3>

A scale factor simply refers to a ratio of two (2) corresponding side lengths or diameter in two (2) similar geometric figures such as triangles.

In this exercise, you're required to construct a dilation of triangle ABC that is centered at B with a scale factor of 2 by using a compass and straightedge.

In conclusion, triangle ABC would remain the same because a dilation generally preserves the measure of an angle.

Read more on scale factor here: brainly.com/question/2826496

#SPJ1

4 0

3 years ago