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Ivan
3 years ago
8

Match the numbers with the correct label

Mathematics
1 answer:
Firdavs [7]3 years ago
4 0

Answer:

a= -3/7

b= -0.2

c = -2/8

Step-by-step explanation:

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-4-3 (6x + 9) = 41?<br> Explain what to do first and what is the answer
lara31 [8.8K]

Answer:

x = -2

Step-by-step explanation:

- 3 x - 6x = -18x

- 4 - 18x + 9 = 41

9 - 4 = 5

- 18x + 5 = 41

- 18x = 41 - 5

- 18x = 36

36 divided by -18 is -2

8 0
3 years ago
To ensure that the horizontal segments are parallel, x must equal
andrezito [222]

Answer:

53

Step-by-step explanation:

Because its 53.

6 0
3 years ago
Simplify using distributive property 4(z+2)+2(3-z)
riadik2000 [5.3K]

Answer:

2z+14

Step-by-step explanation:

you multiply

4*z   =4z

4*2   =8

and then you multiply

2*3=6

2*-z=-2z

the answer is 2z+14

5 0
3 years ago
What is the graph of x-y=1?​
vesna_86 [32]

Answer:

so you have the y intercept of 1 so it is (1,1) bc x=1

Step-by-step explanation:

6 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
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