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dmitriy555 [2]
3 years ago
7

ASAP AND FOR BRAINLIEST 20 POINTSSolve for the variable in the following proportion. 5/25=m25

Mathematics
1 answer:
qaws [65]3 years ago
5 0

Answer:

m = 1 /25 √ 5

Step-by-step explanation:

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Step-by-step explanation: BRAINLIEST PLS

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(PLEASE ANSWER FAST!!!!) Two lines, A and B, are represented by equations given below: Line A: y = x – 2 Line B: y = 3x + 4 Whic
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(-3, -5), because the point satisfies both equations.

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Which is the best buy?
olga55 [171]

Answer:

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Step-by-step explanation:

When you divide each of the prices by the amount of candy bars, B is the lowest. The price for 1 candy bar with A's price is 1.75. The price for 1 candy bar for B's is 1.71. The price for 1 candy bar with C's is 1.73 and the price of 1 candy bar for D's is 1.74. Again, B has the lowest price for 1 candy bar so the correct answer is B.

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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
Consider the line y=5/7x+7. Find the equation of the line that is parallel to this line and passes through the point (-5,-2). Fi
butalik [34]

Step-by-step explanation:

m=5/7

m1=m2=5/7

L;y-(-2)=5/7(x-(-5)

y+2=5/7(x+5)

7y+14=5X+25

7y-5x=11 is parallel lines

m1*m2=-1 is perpendicular lines

m2=-1*7/5

m2=-7/5

L;y+2=-7/5(x+5)

5Y+10=-7x-35

5y+7x=-45 is perpendicular lines

5 0
3 years ago
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