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3 years ago

For f(x)=x^5-x^3 x what is the equation of the tangent line at x=3?

1 answer:

7 0

F(x) = x^5 - x^3 + x
f'(x) = 5x^4 - 3x^2 + 1
f(3) = (3)^5 - (3)^3 + 3 = 243 - 27 + 3 = 219
f'(3) = 5(3)^4 - 3(3)^2 + 1 = 5(81) - 3(9) + 1 = 405 - 27 + 1 = 379

Let the required equation of the tangent line be
y = mx + c; where y = 219, m = 379, x = 3
219 = 379(3) + c = 1,137
c = 219 - 1,137 = -918

Therefore, required equation is
y = 379x - 918

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A line with slope 3 passes through the point (0,10). What is the y-coordinate of the point on the line with x-coordinate 2?

lana [24]

Hello!

So to find the y-coordinate we need an equation for the line. The equation we can use is

y = mx + b

In this we already know m(the slope which is 3) and we can find b by plugging in the points that are given.

10 = 3(0) + b
10 = 0 + b
10 = b

So now that we've found b, we can use this in our equation

y = 4x + b

So now we can plug in the x-coordinate that we know (2) and find the y - coordinate.

y = 4(2) + 10
y = 8 + 10
y = 18

18 is your y-coordinate

4 0

3 years ago

Evaluate 3/5r +5/8swhen r =14 and s =8

Illusion [34]

Step-by-step explanation:

14×5=70

8×8=64

(3/70) + (5/64)

= (3×64)+(5×70)

----‐------------------

70×64

= (192+350)/448

=542/448

= 271/224

5 0

2 years ago

Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round i

dsp73

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s$

Also,

$v=r\omega$

Where

$\omega$ is the angular speed

So,

$\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s$

Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.

8 0

3 years ago

Find the distance between the points (4,10) and (0,0) the answer must be a whole number

Rina8888 [55]

Answer:

d = $2\sqrt{29}$

Step-by-step explanation:

$d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d= \sqrt{(0-4)^2+(0-10)^2}\\d= \sqrt{(-4)^2+(-10)^2}\\d= \sqrt{16+100}\\d= \sqrt{116}\\d=2 \sqrt{29}$

6 0

3 years ago

.98*125<br> 13.75*12<br> 143.43/31

vova2212 [387]

Answer:

n = 6

n, = 6n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2n = 6

n, = 6

n = 0

n, = 0

n = 4

n, = 4

n = 2

n, = 2

3 0

2 years ago