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3 years ago

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F(x)= 2x^3-10x^2+12x-12 Find critical points

1 answer:

elena-s [515]3 years ago

3 0

Answer:

Find where the first derivative is equal to 0

. Enter these values into the original function and simplify to find the critical points.

<h2>(5+√7/3,−284+28√7/27),(5−√7/3,−284−28√7/27)</h2>

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Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y

I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

$\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }$

$\mathbf{ YX = \sqrt{(3)^2+(6)^2} }$

$\mathbf{ YX = \sqrt{9+36} }$

$\mathbf{ YX = \sqrt{45} }$

$\mathbf{ YX = \sqrt{9*5} }$

$\mathbf{ YX = 3 \sqrt{5} }$

Thus; the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

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3 years ago

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The fence encloses at 22 ft2

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3 years ago

Read 2 more answers

Sin4u=2sin 2u cos 2u

yKpoI14uk [10]

Answer:

If you've learnt sin(A+B) = sinAcosB + cosAsinB,

sin(4u)

= sin(2u+2u)

= sin(2u)cos(2u) + cos(2u)sin(2u)

= 2 sin(2u) cos(2u).

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3 years ago

An envelope is 3 inches wide, and it measures 7 inches along the diagonal. What is the envelope’s length? If necessary, round to

Crank

The length of the envelope is 6.3 inches

Step-by-step explanation:

Width of the envelope = 3 inches

Diagonal of the envelope = 7 inches

To find:

The length of the envelope.

Let the length of the envelope be 'l'

We can use pythogoras theorem to calculate the length.

l = √(49-9)

l = √(40)

l = 6.3 inches

The length of the envelope is 6.3 inches

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3 years ago

A dressmaker needs to cut 12-inch pieces of ribbon from rolls of ribbon that are 6 feet in length. How many 12-inch pieces can t

IrinaK [193]

Answer:

60

Step-by-step explanation:

because 12 inches is one foot so there’s 6 feet per 6 feet of ribbon and there’s 10 of these ribbons so 6 times 10 is 60

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3 years ago