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2 years ago

Given f'(x)=(x-4)(6-2x) find the x-coordinate for the relative minimum of the graph f(x).

1 answer:

5 0

The graph you plotted is the graph of f ' (x) and NOT f(x) itself. 

Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get

f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24

So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5

f ' (x) = (x-4)(6-2x)
f ' (3.5) = (3.5-4)(6-2*3.5)
f ' (3.5) = 0.5

The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.

So the final answer is C) 3

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PLEASSSSSE HELPPPPPP This is my third post of this! I really need help

Nesterboy [21]

Answer: x=25

solution:

we know that
The triangle external angle theorem states that: An exterior angle of a triangle is equal to the sum of the opposite interior angles

In this problem:
m∠9=m∠7+m∠8

substitute the given values:
(6x-20)°=30°+4x°

now solve for x:
6x-4x=30+20
2x=50
x=25 which is the final answer:)

6 0

2 years ago

) The density of oil in a circular oil slick on the surface of the ocean at a distance of r meters from the center of the slick

Feliz [49]

Answer:

Therefore the mass of the of the oil is 409.59 kg.

Step-by-step explanation:

Let us consider a circular disk. The inner radius of the disk be r and the outer diameter of the disk be (r+Δr).

The area of the disk

=The area of the outer circle - The area of the inner circle

= $\pi (r+\triangle r)^2- \pi r^2$

$=\pi [r^2+2r\triangle r+(\triangle r)^2-r^2]$

$=\pi [2r\triangle r+(\triangle r)^2]$

Since (Δr)² is very small, So it is ignorable.

∴ $A=2\pi r\triangle r$

The density $\delta (r)= \frac{40}{1+r^2}$

We know,

Mass= Area× density

$=(2r \pi \triangle r)(\frac{40}{1+r^2}})$

Total mass $M=\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i$

Therefore

$\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i=\int_0^5 \frac{80r\pi }{1+r^2}dr$

$=40\pi[ln(1+r^2)]_0^5$

$=40\pi [ln(1+5^2)-ln(1+0^2)]$

$=40\pi ln(26)$

= 409.59 kg (approx)

Therefore the mass of the of the oil is 409.59 kg.

3 0

2 years ago

In order for the parallelogram to be a rhombus x=

Eva8 [605]

Answer:

Step-by-step explanation:

Properties used:-

All sides of rhombus are equal therefore in the triangles

EQUAL SIDE OPPOSITE TO ANGLES, ANGLES BECOME EQUAL

then we use alternate interior angles

Then we get,

3x-11=x+23

2x=34

x=17

4 0

2 years ago

A circle has its center at the center of a square with 12-inch sides. Using 3.14 for pi, find the area of the square not covered

Dmitrij [34]

So first, find the area of the square. 12x12= 144inches^2
Then find the area of the circle. To do this we use pi x r^2. The radius of the circle is 6 inches. Pi x 6^2 = 113.04inches^2.
144 - 113.04 = 30.6, so your answer is 30.6 inches^2

The circles circumference is worked out by using pi x d. The diameter is 12 inches, so 3.14x12 = 37.68 inches.
The squares perimeter is 12 x 4 = 48 inches.
If you round up the circumference of the circle to the nearest integer, you get 38 inches. The ratio would therefore be, 38:48, which can be rounded down to 19:24 :)

5 0

2 years ago

Find the missing side. Round to the nearest 10th.

const2013 [10]

Answer:

5.0

Step-by-step explanation:

Using Pythagorean Theorem:

a² + b² = c²

where a =? b= 12 in and c = 13 in

a² + (12)² = (13)²

a² + 144 = 169

a² = 169 - 144

a² = 25

a = √25 = 5

∴ a = 5.0

Pythagoras Theorem is usually applied when finding one missing side of a right-angle triangle.

4 0

2 years ago