All categories

4 years ago

Find the indicated measurement please

Mathematics

2 answers:

adelina 88 [10]4 years ago

5 0

13.1 is the answer. hope it helps :)

podryga [215]4 years ago

4 0

To find the perimeter, just add all those numbers.

4.8+2.3+2.1+3.9=13.1

Hope this helps!

You might be interested in

What expression is equivalent to ( X 3 ) 4 ?

lianna [129]

Answer:

Step-by-step explanation:

6 0

3 years ago

The area of the rectangular faces of the box are shown . What is the volume of the box

Gnoma [55]

Label the 3 distinct sides of the box. I arbitrarily chose the letters a, b and c.
Use the info about areas as follows:

ab=54 in^2
ac=90 in^2
bc=60 in^2

Here you have 3 equations in 3 unknowns (a, b and c), which is enough info to use to determine a, b and c. Then the volume of the box is a*b*c.

Example: bc = 90, but c = 60/b. You could subst. 60/b for c in the 2nd and 3rd equation, which will eliminate c completely and leave you with 2 equations in 2 unknowns.

Continuing this procedure, I determined that a=9, b=6 and c=10. Thus, the volume of the box is V = 9*6*10 = 540 cubic inches (answer)

3 0

3 years ago

Simplify 3/5(10g-5k)-(-3g+2k)

dybincka [34]

Answer:

9g -5k

Step-by-step explanation:

6 0

4 years ago

I need help with 5 thanks

Mars2501 [29]

For every 3 cups of peanuts there are 2 cups of chocolate

9/3 = 3 so 3*2 = 6

For 9 cups of peanuts there are 6 cups of choclate

3 0

4 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