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3 years ago

Juan Carlos graphs the functions f(x) = 8x - 2 and g(x) = -8x + 2

Mathematics

1 answer:

Mamont248 [21]3 years ago

8 0

Answer:

$x = 0.25$

Step-by-step explanation:

Given

$f(x) = 8x - 2\\ g(x) = -8x + 2$

Required

Determine their point of intersection

To do this, we simply do:

$f(x) = g(x)$

This gives:

$8x - 2 = -8x + 2$

Collect Like Terms

$8x + 8x = 2 + 2$

$16x = 4$

Solve for x

$x = 4/16$

$x = 0.25$

Hence:

Their point of intersection is at $x = 0.25$

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The midpoints of the hexagon are connected to form another hexagon, this pattern continues indefinitely. If the area of the 99th

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7 0

2 years ago

Read 2 more answers

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)

vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

$y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\$

Apply substitution to get

$y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right) + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x) + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x) + \left(a - 6b\right)\cos(x) = \sin(x)\\\\$

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0

2 years ago

~please only answer if you know for sure~

NARA [144]

JKL is similar to PQR

so they want PR

JKL is similar to PQR

So PR is similar to JL

And they also give us the sides KL and QR

JKL is similar to PQR

Both of those sides are similar, so we can form a proportion.

$\sf { \frac{PR}{JL} = \frac{QR}{KL} }$

Plug in the numbers:

$\sf { \frac{PR}{6} = \frac{15}{9} }$

Cross multiply:

$\sf{ PR \times 9 = 6 \times 15}$

Isolate PR

$\sf{ PR = \frac{6 \times 15}{9}}$

Simplify it

$\sf{ PR = 10}$

So your final answer is

$\boxed{\bf{10 centimeters}}$

6 0

3 years ago

Read 2 more answers