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2 years ago

Solve x2 + 10x = 24 by completing the square. Which is the solution set of the equation?

Mathematics

1 answer:

Lilit [14]2 years ago

4 0

Answer:

{ - 12, 2 }

Step-by-step explanation:

Given

x² + 10x = 24

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(5)x + 25 = 24 + 25

(x + 5)² = 49 ( take the square root of both sides )

x + 5 = ± $\sqrt{49}$ = ± 7 ( subtract 5 from both sides )

x = - 5 ± 7

Thus

x = - 5 - 7 = - 12

x = - 5 + 7 = 2

Solution set is { - 12, 2 }

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This question has two parts. Be sure to answer both parts of the question. One inequality in a system is shown below. -12x + 3y

fgiga [73]

The inequality is -12x + 3y > 9.

PART A:

The sytem has no solution if inequality does not share a common area. The inequality -12x + 3y > 9 consist the region to left of line -12x + 3y = 9. So for no solution the region to left of equation -12x + 3y = 9 is suitable.

Thus inequality for no solution is, -12x + 3y < 9.

PART B:

For infinite solution the region of both inequality must overlap each other, or the inequality is same with some multiplication of divison factor. So inequality for infinite many solutions is,

$\begin{gathered} (-12x+3y>9)\times-1 \\ 12x-3y$

Thus inequality for infinite many solution is 12x - 3y < -9.

7 0

1 year ago

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

Roman55 [17]

Answer:

Standard error (S.E) = 0.048 ≅ 0.05

Step-by-step explanation:

<u>Explanation</u>:-

Given sample size 'n'=76

Given the dean who randomly selects 76 students and finds that 58 out of 76 are receiving financial aid.

Sample proportion

$p = \frac{x}{n} = \frac{58}{76} = 0.7631$

The standard error is determined by

$S.E = \sqrt{\frac{p(1-p)}{n} }$

$S.E = \sqrt{\frac{0.763(0.2369}{76} }$

Standard error is 0.048 ≅ 0.05

3 0

3 years ago

Please help and explain why, so that I understand.

chubhunter [2.5K]

Hello there!

The number of people in different cities should be represented on a graph as Two-dimensional.

I'll explain why: The number of dimensions you need on a graph is equal to the number of variables that you are studying. In this case, you have two variables in the study:

1)The number of people
2)The city where they live

You'll need a two-dimensional graph with two axes in which one of the axes represents the number of people and the other axis represents the city. If you were also studying the age of those people, you'll need one more dimension to include this new variable.

Have a nice day!

8 0

3 years ago

Suppose that we have two identical boxes: box 1 and box 2. box 1 contains 5 red balls and 3 blue balls. box 2 contains 2 red bal

sineoko [7]

(a) The probability of drawing a blue marble at random from a given box is the number of blue marbles divided by the total number of marbles. We assume that the probability of selecting one of two boxes at random is 1/2 for each box.

... P(blue) = P(blue | box1)·P(box1) + P(blue | box2)·P(box2) = (3/8)·(1/2) + (4/6)·(1/2)

... P(blue) = 25/48 . . . . probability the ball is blue

(b) P(box1 | blue) = P(blue & box1)/P(blue) = (P(blue | box1)·P(box1)/P(blue)

... = ((3/8)·(1/2))/(25/48)

... P(box1 | blue) = 9/25 . . . . probability a blue ball is from box 1

8 0

3 years ago

A base of ten raised to a negative exponent corresponds to a number that is-

Masteriza [31]

Answer:

A base of ten raised to a negative exponent corresponds to a number that is between 0 and 1.

Step-by-step explanation:

A base of ten raised to a negative exponent, say - 1 will correspond to the number, $10^{- 1} = \frac{1}{10^{1}} = \frac{1}{10}$ .

Again, a base of ten raised to a negative exponent, say - 2 will correspond to the number, $10^{- 2} = \frac{1}{10^{2}} = \frac{1}{100}$ .

{Since we know the property of exponents as $x^{- a} = \frac{1}{x^{a}}$ }

Therefore, all those numbers are between 0 and 1.

Hence, a base of ten raised to a negative exponent corresponds to a number that is between 0 and 1. (Answer)

3 0

3 years ago