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3 years ago

Given the quadratic equation 8-6x=-2x^2 complete the following:

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

a. $2x^2-6x+8$

b. a=2, b=-6, and c=8

Step-by-step explanation:

The standard form of a quadratic is $ax^2+bx+c$ written in descending order and can be found by using inverse operations to rearrange terms.

$8-6x=-2x^2\\\\8-6x+2x^2=-2x^2+2x^2\\\\2x^2-6x+8=0$

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(X^3+1)dividend (x-1)

bazaltina [42]

$x^3=x\cdot x^2$ and $x^2(x-1)=x^3-x^2$ . So we have a remainder of

$(x^3+1)-(x^3-x^2)=x^2+1$

$x^2=x\cdot x$ and $x(x-1)=x^2-x$ . Subtracting this from the previous remainder gives a new remainder

$(x^2+1)-(x^2-x)=x+1$

$x=x\cdot1$ and $1(x-1)=x-1$ . Subtracting this from the previous remainder gives a new one of

$(x+1)-(x-1)=2$

and we're done since 2 does not divide $x$ . So we have

$\dfrac{x^3+1}{x-1}=x^2+x+1+\dfrac2{x-1}$

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3 years ago

Alexa is planning to drive across the United States in

ivann1987 [24]

Answer:

a=112

b=6

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Step-by-step explanation:

you just plug the number into the equation

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3 years ago

Read 2 more answers

X = A 21.6 B 12.9 C 9.1

Artist 52 [7]

Answer:

B: 12.9

Step-by-step explanation:

If A B and C are the alternatives, the answer is B: 12.9

The reason is that the hypotenuse can't be smaller than one of its sides, that eliminates alternative C.

And also, the hypotenuse can't be bigger than segments AC + BC, since the side AC is bigger and AC measures 10.8, the hypotenuse would have to measure < 21.6

Any questions, comment.

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3 years ago

Read 2 more answers

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th

OlgaM077 [116]

Answer:

Interval [16.34 , 21.43]

Step-by-step explanation:

First step. Calculate the mean

$\bar X=\frac{(23+18+23+12+13+23)}{6}=18.666$

Second step. Calculate the standard deviation

$\sigma =\sqrt{\frac{(23-18.666)^2+(18-18.666)^2+(23-18.666)^2+(12-18.666)^2+(13-18.666)^2+(23-18.666)^2}{6}}$

$\sigma=\sqrt\frac{18.783+0.443+18.783+44.435+5.666+18.783}{6}$

$\sigma=\sqrt{17.815}=4.22$

As the number of data is less than 30, we must use the t-table to find the interval of confidence.

We have 6 observations, our level of confidence DF is then 6-1=5 and we want our area A to be 80% (0.08).

We must then choose t = 1.476 (see attachment)

Now, we use the formula that gives us the end points of the required interval

$\bar X \pm t\frac{\sigma}{\sqrt n}$

where n is the number of observations.

The extremes of the interval are then, rounded to the nearest hundreth, 16.34 and 21.43

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3 years ago

X^3+6x^2-17x+2-x^3-x^2-11x+36

Rudiy27

X^3+6x^2-17x+2-x^3-x^2-11x+36 = 5x^2 - 28x + 38

4 0

3 years ago