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Alona [7]
3 years ago
15

HELPPP ILL DO ANYTHING HELPPPP

Mathematics
1 answer:
katovenus [111]3 years ago
3 0

Answer:

-2

Step-by-step explanation:

hi! to find f(-3) on this graph, go to the x-value -3 and find the y-value there. when we look at -3 on the x-axis, we go down and see that there is a point at (-3,-2). therefore, f(-3)=-2.

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Look at the figure rectangle PQRS, find m angel P
Aliun [14]
144 degrees, q=s, so 3a=4a-12. a=12, and 12*12= 144
5 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
I need help quick please
qwelly [4]
A=90
B=53
The reason I am saying this is because you can see the 90 degree angle in A
8 0
2 years ago
Read 2 more answers
Please help. Its my last, please show at least some steps as I need to show work. Tysm
Fofino [41]
What’s your question :D
5 0
3 years ago
2x +4 (3 - 2x) = 3(2x + 2)/6 + 4<br> from my math book
liberstina [14]

Answer:

x = 1

Step-by-step explanation:

2x +4 (3 - 2x) = 3(2x + 2)/6 + 4

2x+12-8x = (6x+6)/6 +4

-6x + 12 = x+1+4

12-5 = 7x

7x = 7

x=1

Hope this helps!

4 0
2 years ago
Read 2 more answers
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