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3 years ago

15

Can you guys help me ?

1 answer:

pashok25 [27]3 years ago

6 0

The graph should look like this
Pattern for each point is up 2 over 3 because the slope is 2/3

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The function = 3describes the volume of a cube, , in cubic inches, whose length, width, and height each measure inches. Find the

kolezko [41]

The correct structure of the question is as follows:

The function f(x) = x^3 describes a cube's volume, f(x) in cubic inches, whose length, width, and height each measures x inches. If x is changing, find the (instantaneous) rate of change of the volume with respect to x at the moment when x = 3 inches.

Answer:

Step-by-step explanation:

Given that:

f(x) = x^3

Then;

V = x^3

The rate whereby V is changing with respect to time is can be determined by taking the differentiation of V

dV/dx = 3x^2

Now, at the moment when x = 3;

dV/dx = 3(3)^2

dV/dx = 3(9)

dV/dx = 27 cubic inch per inch

Suppose it is at the moment when x = 9

Then;

dV/dx = 3(9)^2

dV/dx = 3(81)

dV/dx = 243 cubic inch per inch

6 0

3 years ago

23,681,714 rounded to the nearest million

Katena32 [7]

24,000,000 because 600,000 is more than 1/2 1,000,000 so you round up.

5 0

3 years ago

John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21 About how far does

Agata [3.3K]

Answer:

John ski down the mountain is 1285.37 feet.

Step-by-step explanation:

Given : John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21.

To find : About how far does John ski down the mountain ?

Solution :

We draw a rough image of the question for easier understanding.

Refer the attached figure below.

According to question,

Let AB be the height of mountain i.e. AB=1200 feet

The angle of depression is 21 i.e. $\theta=21^\circ$

We have to find how far does John ski down the mountain i.e. AC = ?

Using trigonometric,

$\cos\theta = \frac{AB}{AC}$

$\cos(21)= \frac{1200}{AC}$

$AC=\frac{1200}{\cos(21)}$

$AC=1285.37$

Therefore, John ski down the mountain is 1285.37 feet.

7 0

4 years ago

Second pictures of my last question

Elanso [62]

Answer:

you really did not give us the full pictuer

Step-by-step explanation:

8 0

3 years ago

I have no idea how to do this, it is due in two days. Hopefully someone sees this before then.

elena-14-01-66 [18.8K]

Hello,

$m\ \widehat{ABC}=x\\m\ \widehat{BAC}=2*x\\\\So:\\ x+2x=90^o\\x=30^o\\$

$cos(30^o)=\dfrac{\sqrt{3} }{2} \\$

In the triangle ABC,

$cos(30^o)=\frac{BC}{BA} \\\\BA=\dfrac{cos(30^o)}{BC} \\\\BA=\frac{\dfrac{\sqrt{3} }{2} }{24} =16*\sqrt{3} \\\\$

$sin(30^o)=\dfrac{1 }{2} =\dfrac{AC}{AB} \\\\AC=\dfrac{1}{2} *16\sqrt{3} =8\sqrt{3}$

In the triangle ACB,

$cos(30^o)=\dfrac{AC}{AL} \\\\AL=\dfrac{8\sqrt{3} *2}{\sqrt{3} } =16\\$

6 0

3 years ago