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Hoochie [10]
3 years ago
14

Write the equation of the line that passes through (−3, 5) and (2, 10) in slope-intercept form.

Mathematics
1 answer:
Paladinen [302]3 years ago
7 0

Answer:

gradient =y2-y1÷x2-x1

=10-5÷2-(-3)

1

y-y1=m(x-x1)

y-5=1(x-(-3))

y=x+2

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What is the same as 1.5 in a fraction
jeka94

Answer:

the answers is 3 over 2 in fraction form

4 0
2 years ago
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Serjik [45]

Answer

B

explanation

3480ones = 3480x1=3480

8 0
1 year ago
Write the equation of a line that passes through the points ( 2 , - 5 ) and ( 8 , - 2 ).
Zielflug [23.3K]

Answer:

y = 1/2x - 6

Step-by-step explanation:

y2 - y1 / x2 - x1

-2 - (-5) / 8 - 2

3 / 6

= 1/2

y = 1/2x + b

-5 = 1/2(2) + b

-5 = 1 + b

-6 = b

3 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
What values of k will make the line containing points (k,3) and (-2,1) parallel to the line through (5, K) and (1,0)
rjkz [21]

Answer:

k = -4 or 2

Step-by-step explanation:

For the lines to be parallel, the lines would need to be the same slope. This means to find the slope of each we would take the difference of the rise over the run.

\frac{1-3}{-2-k}=\frac{-2}{-2-k}

and

\frac{0-k}{1-5}=\frac{-k}{-4}

To find K, set them equal.

\frac{-2}{-2-k} = \frac{-k}{-4}\\-k(-2-k) = -2*-4\\2k+k^2=8\\k^2+2k-8=0

Factor the quadratic and solve for K.

(k+4)(k-2)=0

k+4=0 so k=-4

k-2=0 so k=2

4 0
3 years ago
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