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AURORKA [14]
3 years ago
14

HELP! Please and thanks

Mathematics
1 answer:
melisa1 [442]3 years ago
5 0

I think it is c 92                             [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

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Use the Pythagorean identity on page 10 with x = 4 and y= 3 to generate a Pythagorean triple​
Dmitry [639]

Answer:

  c.  7, 24, 25

Step-by-step explanation:

For the formulas ...

  • a=x^2-y^2
  • b=2xy
  • c=x^2+y^2

we can use the given values of x and y to find the corresponding Pythagorean triple:

  a = 4^2 -3^2 = 16 -9 = 7

  b = 2·4·3 = 24

  c = 4^2 +3^2 = 16 +9 = 25

The generated Pythagorean triple is (7, 24, 25), matching answer C.

6 0
2 years ago
I need this ASAP Solve: 2x-10/4= 3x
Dovator [93]
X= -5/2 (0.4)

or

x= -2 1/2 (2.5)
4 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
2 years ago
Consider the sphere enclosed by the cylinder. If the diameter of the sphere is 12 cm, what is the volume of the cylinder? A) 36π
kogti [31]
To find the volume of a cylinder , you will use the formula
V = Bh, where B is the area of the circular base.
Pi x r^2 x h
pi x 6^2 x 12
V = 432
The volume of the cylinder is 432pi cm³.
6 0
3 years ago
Read 2 more answers
What is the inverse function of f(x) = -2?<br> Hurry please
notsponge [240]

Answer:

A

Step-by-step explanation:

2x/x-1

4 0
3 years ago
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