Simplify fully 4x + 3y -2 + 2x - 8y -6

Answers

Answer 1

Answer: Hi, Amyleighx!

First we want to identify any common factors.
4x+3y-2+2x-8y-6
3y-8y=-5y
4x+2x=6x
-2-6=-8
-5y+6x-8

I hope this helps;)

Answer 2

Answer: combine like terms. you can combine 4x and 2x, so that is 6x. now combine 3y and-8y and get -5y. combine-2 and -6 and get -8. now put ur terms together. 6x-5y-8

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A snack cart sells lemonade for $2 and hot dogs for &5. The vendor sold 86 items today for a total of $330. Which equation is true?

Answers

2x+5x=330. I'm pretty sure that's it but you didn't list any answer choices.

I agree as well with both these people if you had listed your options we would have probably a better chance of getting the answer right but I do belive that the equation is right

Farmingdale is located 8 mi due west of Cambridge. Bayville is located due east of Cambridge, and Louisville is located due north of Cambridge. Louisville is 17 mi from Farmingdale and 24 mi from Bayville.How far is Bayville from Cambridge? Round to the nearest tenth of a mile.

Answers

Bayville is 18.7 mi from Cambridge.

152/26 rounded to three decimal places

Answers

Answer:

6.000

Step-by-step explanation:

156/26=6

to 3d.p = 6.000

Which of the following functions are solutions of the differential equation y'' + y = sin(x)?a) y= sinxb) y= cosx
c) y=1/2sinx
d) -1/2xcosx

Answers

Answer:

Option (d)

Step-by-step explanation:

Given,

y" +y=sin x ...........(1)

The particular solution

$y_p=A x sinx +Bx cosx$

$y'_p=Axcosx+Asinx+B cosx-Bxsinx$

$y$

Putting the value of y" and y in equation (1)

$2Acosx-Axsinx-2Bsinx-Bxcosx+Axsinx+Bxcosx = sinx$

$\Rightarrow 2Acosx-2Bsinx=sinx$

Therefore 2A =0 -2B=1

⇒A=0 $\rightarrow B=-(1)/(2)$

Therefore $y_p=-(1)/(2) x cosx$

Final answer:

The solutions of the differential equation y'' + y = sin(x) are y = cos(x), y = (1/2)sin(x), and y = -(1/2)xcos(x).

Explanation:

To determine which of the given functions are solutions of the differential equation y'' + y = sin(x), we can substitute each function into the equation and check if it satisfies the equation. Let's go through each option:

Substituting y = sin(x) into the equation, we get -sin(x) + sin(x) = sin(x), which is not true. So, y = sin(x) is not a solution.
Substituting y = cos(x) into the equation, we get -cos(x) + cos(x) = sin(x), which is true. So, y = cos(x) is a solution.
Substituting y = (1/2)sin(x) into the equation, we get -(1/2)sin(x) + (1/2)sin(x) = sin(x), which is true. So, y = (1/2)sin(x) is a solution.
Substituting y = -(1/2)xcos(x) into the equation, we get (-1/2)xcos(x) + (1/2)xcos(x) = sin(x), which is true. So, y = -(1/2)xcos(x) is a solution.

Therefore, the solutions of the differential equation y'' + y = sin(x) are y = cos(x), y = (1/2)sin(x), and y = -(1/2)xcos(x).

Learn more about Solutions of a differential equation here:

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Suppose we toss a coin 4 times. how many different sequences of outcomes are possible if order matters? (for example, hhtt is one outcome and htht is counted as a different outcome)

Answers

Solution: Each coin toss has 2 possible outcomes "Head" and "Tail". So if we flip a coin four times, the number of possible outcomes are:

$2^(4) = 16$ outcomes.

Let H denotes the Head and T denotes the Tail, then the 16 possible outcomes are enumerated below:

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,

THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT

65 Points - I have class in 1hr and I still can't figure this out... pls give full explanation and correct answer or else you'll be reported for "point greediness" --- thx ya'llAB is tangent to the circle k(O) at B, and ADis a secant, which goes through O. Point O is between A and D∈k(O). Find m∠BAD and m∠ADB, if measure of arc BD is 110°20'.

Answers

Sounds like the situation in the sketch below...

There is a theorem that says the angle (BAD) formed by a tangent (AB) and a secant (AD) to a circle (O) is half the difference of the intercepted arcs (BD and BD'):

$m\angle BAD=\frac{110^\circ20'-69^\circ40'}2=40^\circ40'$

Triangle BAD is a right triangle, so

$m\angle ADB=90^\circ-m\angle BAD=49^\circ20'$

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