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katrin [286]
3 years ago
7

Help me Please !!!! You will receive Brainliest!!!

Mathematics
1 answer:
Gnesinka [82]3 years ago
7 0
(-3, -3) and (2, 7).

The solutions to a graphed system of equations are where the two graphs cross each other. So in this case the two equations cross at (-3, -3) and (2, 7), and those points are the solutions to the system of equations.
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A diver went 25.65 feet below the surface of the ocean, and then 16.5 feet further down, he rose 12.45 feet.
Rasek [7]

If you are trying to find where he is the answer is -29.7

Hope this helps!


5 0
3 years ago
Read 2 more answers
Solve the equation. Check your answer:<br><br> 14+6a-8=18
Ivanshal [37]
A=2...
Subtract 14 from both sides,then subtract 8 from both sides. Your left with
6a=12
6x1=6
6x2=12
A=2
5 0
3 years ago
Express 0.001 in scientific notation by counting decimal places
kati45 [8]
1 x 10-3 make sure the -3 is above the 10.
7 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
I need help ASAP!!<br>Mathematics​
Ilia_Sergeevich [38]

Answer:AGC and DGH are opposite angles.

Step-by-step explanation:

So, Y+10=3x-40

Y-3x=-50-(1)

3x-40=2x

3x-2x=40

x=40

Y-3x40=-50

y-120=-50

y=-50+120

y=70

Hence, the answer is A 70°

3 0
2 years ago
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