A year on mars is 1.88 times as long as a year on earth .An earth year lasts 365.3 days.Find the length of a year on mars

An ice cream cone has a diameter of 3 inches.The distance from the top of the cone to the point at the bottom (height) is 5 inch

solong [7]

Answer:

11.781 cubic inches (round accordingly)

Step-by-step explanation:

The formula for the volume of a cone is:

$V=\frac{1}{3}\pi r^{2} h$

where r is the radius and h is the height.

For our ice cream cone:

d = 3 inches

r = 1.5 inches (the radius is half of the diameter)

h= 5 inches

So all we have to do is plug in our values and solve for the volume:

$\frac{1}{3}\pi (1.5)^{2}(5) = 11.781$

So our final answer is around 11.8 cubic inches.

3 0

3 years ago

A rectangular solid with a square base has a volume of 4096 cubic inches - determine the dimensions that yield the minimum surfa

Bumek [7]

Answer:

side of base, a = 10.1 inches, height, h = 40.1 inches

Step-by-step explanation:

Volume of rectangular solid, V = 4096 cubic inches

Let the side of base is a and the height is h.

$V = a^2h\\\\4096 = a^2 h ..... (1)$

surface area of the solid

$S = 2a^2 + 4 ah \\\\S = 2a^2 + 4 \times a \times \frac{4096}{a^2} from (1)\\\S = 2a^2 + \frac{4096}{a}\\\\\frac{dS}{da}= 4 a - \frac{4096}{a^2}\\\\4 a - \frac{4096}{a^2} = 0 \\\\a^3 = 1024\\\\a =10.1 inches$

So, h = 40.2 inches

5 0

3 years ago

what ordered pair is 2 spaces to the left and 4 spaces down from the point graphed on the coordinate plane?

kykrilka [37]

C. (-1, -2) would be your answer
That transformation is just a translation (x-2, y-4) from point (1,2)

8 0

3 years ago

the surface area of earth is about 196.9 million square miles. The land area is about 57.5 million square miles, and the rest is

kow [346]

Answer: 29.2%

Step-by-step explanation:

The total surface area is 196.9 million square miles and out of this, land occupies 57.5 million square miles.

The probability that a meteorite would hit land is the percentage of and out of the entire planet area:

= Land area/ Total earth surface area

= 57.5/ 196.9

= 29.2%

8 0

3 years ago

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

3 years ago