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OlgaM077 [116]
3 years ago
8

How many cubic inches of storage space are in Leo’s cabinet ? Please help :)

Mathematics
1 answer:
Paha777 [63]3 years ago
4 0

Answer:

the amount of space present is 15,120 cubic inches

Step-by-step explanation:

To get the amount of space, we simply calculate the volume of the space

This is the area of the base multiplied by the length

Mathematically, we have this as;

Area of the triangle multiplied by the length of the cabinet

We have this as;

(1/2 * 20 * 21) * 72

= 210 * 72

= 15,120 cubic inches

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A bug is moving back and forth on a straight path. The velocity of the bug is given by vt=t2-3t. Find the average acceleration o
g100num [7]

Answer:

2 unit/time²

Step-by-step explanation:

Given the equation:

v(t) =t^2-3t

At interval ; 1, 4

V(1) = 1^2 - 3(1)

V(1) = 1 - 3

V(1) = - 2

At t = 4

V(4) = 4^2 - 3(4)

V(4) = 16 - 12

V(4) = 4

Average acceleration : (final - Initial Velocity) / change in time

Average acceleration = (4 - (-2)) ÷ (4 - 1)

Average acceleration = (4 + 2) / 3

Average acceleration = 6 /3

Average acceleration = 2

3 0
3 years ago
Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

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6 0
3 years ago
Help? Please?<br> I really suck at geometry
Maslowich

Answer:

x = 14

y = 4

Explanation:

Ok so, just from looking at the two triangles i can tell they're congruent right triangles. I used different colors to show which sides of the triangle correspond and are equal to each other in my attatched photo.

So the side thats equal to x is the same length as the side that's equal to y+10 on the other triangle.

So we can write the equation x = y + 10.

Using this same method, the side that's equal to x + 2 is the same length as the side that's equal to 4y on the other triangle.

So, we can write the equation 4y = x + 2.

Now we have the equations \left \{ {{x=y+10} \atop {4y=x+2}} \right. you could rewrite to be in slope- intercept form so they're easier to graph. But a graphing calculator online would plot it just fine.

If you graph these two equations they'll intersect at the solution ( 14, 4 ). I'll include the graph in my images as well.

To check your answer, you can plug in x and y and see if the triangle sides end up being the same length. I did and it was correct.

8 0
3 years ago
HELP!!!<br> ANSWER FAST. PLEASEEE
kari74 [83]

Answer:

The answer would be $2.

Have a nice day.

4 0
3 years ago
Read 2 more answers
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
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