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3 years ago

Solve for x, x^2-3x-5=0

1 answer:

7 0

Okay so for this one, we have to isolate our X.Lets start by moving the negative 5 on the other side of the equal sign. Remember to not mix a number that is with an X with a normal number (in this case it would be the -3x and the -5). So in putting our -5 on the other sign it becomes positive. Here’s what we have: x^2-3x=5. x^2 is the same thing as 2x.So now you have to minus 3x off of your 2x. This gives us -1x, but can just be put as -x. So now we have -x=5. You’re answer could either be 5 or -5 if you decided to divide it by -1 seeing as your -x is technically -1.

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1) 225 What is the square root of 225 simplified?

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Answer: 15

Step-by-step explanation: This is one of those square roots you'll memorize really quickly. You don't even have to simplify it, since it's an integer.

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2 years ago

At Elmwood Middle School there are 4 teachers for every 68 students. There are 425 students enrolled at the school. How many tea

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Answer:

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Step-by-step explanation:

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2 years ago

Which factor do 64x^2 + 48x + 9 and 64x^2 - 9 have in common?

Ber [7]

Start with the second expression. That factors easily into (8x-3)(8x+3).
Since 64x^2 + 48x + 9 has no negative signs, I'd bet that the common factor is (8x+3). Divide 64x^2 + 48x + 9 by that to verify.

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3 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

Read 2 more answers

OB is the bisector of AD and EC. . EO = 2x – 20 and OC = 70 – x. Solve for x.

mixas84 [53]

We know that OB bisects EC where they intersect would be at point O,
so the two lengths EO and OC must be equal.

2x - 20 = 70 - x

[adding x + 20 to each side]

2x - 20 + x + 20 = 70 - x + x + 20

[simplifying]

3x = 90

[dividing each side by 3]

x = 30 would be the answer

4 0

2 years ago