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3 years ago

This system has one solution, what is the y-coordinate of the solution.

Mathematics

1 answer:

Lyrx [107]3 years ago

5 0

The coordinates would be -3, 10.
This Website helps a lot. www.meta-calculator.com

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How many dogs were household pets in the U.S in 2012, in scientific notation

Irina18 [472]

Answer:

Step-by-step explanation:

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3 years ago

A ball is draw at random form a box containing 6 red balls, 4 white balls and 5 blue balls. what is the probability that the bal

makkiz [27]

1 = 6 out of 15
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3 0

3 years ago

Solve the linear equation. 3x = 24 x = ?

trapecia [35]

Answer:

3x =24

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3 years ago

Are 35° and -685° coterminal?

Naya [18.7K]

Answer: YES

<u>Step-by-step explanation:</u>

Coterminal means they are located at the same place on the Unit Circle but are n-rotations clockwise or counterclockwise.

Start with 35° and continue to subtract 360° (which is one rotation) until you reach -685° or pass it. If you reach -685°, then it is coterminal. If you pass it, then it is not coterminal.

35° - 360° = -325°

-325° - 360° = -685°

So, -685° is 2 rotations counterclockwise to 35°, which means the two angles are coterminal to each other.

6 0

3 years ago

family of solutions of the second-order DE y y 0. Find a solution of the second-order IVP consisting of this differential equati

Anna35 [415]

Answer:

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$

Step-by-step explanation:

Given

$y=c_1e^x +c_2e^{-x$

$y(1) = 0$

$y'(1) =e$

Required

The solution

Differentiate $y=c_1e^x +c_2e^{-x$

$y' = c_1e^x - c_2e^{-x}$

Next, we solve for c1 and c2

$y(1) = 0$ implies that; x = 1 and y = 0

So, we have:

$y=c_1e^x +c_2e^{-x$

$0 = c_1 * e^1 + c_2 * e^{-1}$

$0 = c_1 e + \frac{1}{e}c_2$ --- (1)

$y'(1) =e$ implies that: x = 1 and y' = e

So, we have:

$y' = c_1e^x - c_2e^{-x}$

$e = c_1 * e^1 - c_2 * e^{-1}$

$e = c_1 e - \frac{1}{e}c_2$ --- (2)

Add (1) and (2)

$0 + e = c_1e + c_1e + \frac{1}{e}c_2 - \frac{1}{e}c_2$

$e = 2c_1e$

Divide both sided by e

$1 = 2c_1$

Divide both sides by 2

$c_1 = \frac{1}{2}$

Substitute $c_1 = \frac{1}{2}$ in $0 = c_1 e + \frac{1}{e}c_2$

$0 = \frac{1}{2} e+ \frac{1}{e}c_2$

Rewrite as:

$\frac{1}{e}c_2 = -\frac{1}{2} e$

Multiply both sides by e

$c_2 = -\frac{1}{2} e^2$

So, we have:

$y=c_1e^x +c_2e^{-x$

$y = \frac{1}{2}e^x -\frac{1}{2} e^2 * e^{-x}$

$y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}$

6 0

3 years ago