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3 years ago

Which two number are 5 unit away from 0 on the number line

Mathematics

1 answer:

geniusboy [140]3 years ago

3 0

Answer:

<h2>You've worked with numbers on a number line. You know how to graph numbers like 0, 1, 2, 3, etc. on the number line. There are other kinds of numbers that can be graphed on the number line, too. Let's see what they look like and where they are located on the number line.</h2>

Step-by-step explanation:

<h2>Hope this helps!!</h2>

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A Test subject is randomly selected for a diabetes test. What is the probability of getting a subject who is not diabetic, given

zheka24 [161]

Probability of getting a subject that is not diabetic, given that the result is negative is 90.3%

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2 years ago

Classify Angle 5 & Angle 6 by choosing the correct term from the choices below.

vampirchik [111]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

In the given figure, the angles 5 and 6 form Vertical opposite angle pair.

so, correct choice is Vertical

6 0

2 years ago

A perpendicular bisector, , is drawn through point C on .

Liula [17]

Answer:

The x-intercept of CD is B(18/5,0). The point C(32,-71) lies on the line CD.

Step-by-step explanation:

the x-intercept of CD is[ A(3,0) B(18/5,0) C(9,0) D(45/2,0) ] . Point [ A(-52,117) B(-20,57) C(32,-71) D(-54,-128) ] lies on CD.

Given :

CD is perpendicular bisector of AB.

The coordinates of point A are (-3, 2) and the coordinates of point B are (7, 6).

C is the midpoint of AB.

$C=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{7-3}{2},\frac{2+6}{2})=(2,4)$

The coordinates of C are (2,4).

Line AB has a slope of: $m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{7-(-3)}=\frac{4}{10}=\frac{2}{5}$

The product of slopes of two perpendicular lines is -1. Since the line CD is perpendicular to AB, therefore the slope of CD : $m_2=-\frac{5}{2}$

The point slope form of a line is given by:

$y-y_1=m(x-x_1)$

The slope of line CD is $-\frac{5}{2}$ and the line passing through the point (2,4), the equation of line CD can be written as:

$y-4=-\frac{5}{2}(x-2)\\y=-\frac{5}{2}x+5+4\\y=-\frac{5}{2}x+9 .... (1)$

The equation of CD is $y=-\frac{5}{2}x+9$

In order to find the x-intercept, put y=0.

$0=-\frac{5}{2}x+9\\\frac{5}{2}x=9\\x=\frac{18}{5}$

Therefore the x-intercept of CD is B(18/5,0).

Put x=-52 in eq(1).

$y=-\frac{5}{2}(-52)+9=139$

Put x=-20 in eq(1).

$y=-\frac{5}{2}(-20)+9=59$

Put x=32 in eq(1)

$y=-\frac{5}{2}(32)+9=-71$

Put x=-54 in eq1).

$y=-\frac{5}{2}(-54)+9=144$

Thus, only point (32,-71) satisfies the equation of CD. Therefore the point C(32,-71) lies on the line CD.

8 0

3 years ago

Mika is a local farmer who is interested in how often residents in her town go to a farmer's market each month. She surveys 124

TEA [102]

Answer:

The margin of error is of 0.3012, and it means that we should be 99% confident that the population mean would be within 0.3012 of the sample mean.

Step-by-step explanation:

Margin of error

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation and n is the size of the sample.

Standard deviation of 1.3

This means that $\sigma = 1.3$

She surveys 124 families

This means that $n = 124$

Margin of error and meaning:

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 2.58\frac{1.3}{\sqrt{124}}$

$M = 0.3012$

The margin of error is of 0.3012, and it means that we should be 99% confident that the population mean would be within 0.3012 of the sample mean.

6 0

2 years ago

Hitunglah nilai x ( jika ada ) yang memenuhi persamaan nilai mutlak berikut . Jika tidak ada nilai x yang memenuhi , berikan ala

Julli [10]

(a). The solutions are 0 and ⁸/₃.

(b). The solutions are 1 and ¹³/₃.

(c). The equation has no solution.

(d). The only solution is ²¹/₂₀.

(e). The equation has no solution.

<h3>Further explanation</h3>

These are the problems with the absolute value of a function.

For all real numbers x,

$\boxed{ \ |f(x)|=\left \{ {{f(x), for \ f(x) \geq 0} \atop {-f(x), for \ f(x) < 0}} \right. \ }$

<u>Problem (a)</u>

|4 – 3x| = |-4|

|4 – 3x| = 4

$\boxed{ \ 4 - 3x \geq 0 \ } \rightarrow \boxed{ \ 4\geq 3x \ } \rightarrow \boxed{ \ x\leq \frac{4}{3} \ }$

For 4 – 3x = 4

Subtract both sides by four.

-3x = 0

Divide both sides by -3.

x = 0

Since $\boxed{ \ 0\leq \frac{4}{3} \ }$ , x = 0 is a solution.

$\boxed{ \ 4 - 3x < 0 \ } \rightarrow \boxed{ \ 4 < 3x \ } \rightarrow \boxed{ \ x > \frac{4}{3} \ }$

For -(4 – 3x) = 4

-4 + 3x = 4

Add both sides by four.

3x = 8

Divide both sides by three.

$x = \frac{8}{3}$

Since $\boxed{ \ \frac{8}{3} > \frac{4}{3} \ }$ , $\boxed{ \ x = \frac{8}{3} \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 0 \ and \ \frac{8}{3} \ }$

————————

<u>Problem (b)</u>

2|3x - 8| = 10

Divide both sides by two.

|3x - 8| = 5

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x - 8 = 5

Add both sides by eight.

3x = 13

Divide both sides by three.

$x = \frac{13}{3}$

Since $\boxed{ \ \frac{13}{3} \geq \frac{4}{3} \ }$ , $\boxed{ \ x = \frac{13}{3} \ }$ is a solution.

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x – 8) = 5

-3x + 8 = 5

Subtract both sides by eight.

-3x = -3

Divide both sides by -3.

x = 1

Since $\boxed{ \ 1 < \frac{8}{3} \ }$ , $\boxed{ \ x = 1 \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 1 \ and \ \frac{13}{3} \ }$

————————

<u>Problem (c)</u>

2x + |3x - 8| = -4

Subtracting both sides by 2x.

|3x - 8| = -2x – 4

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x – 8 = -2x – 4

3x + 2x = 8 – 4

5x = 4

$x = \frac{4}{5}$

Since $\boxed{ \ \frac{4}{5} \ngeq \frac{8}{3} \ }$ , $\boxed{ \ x = \frac{4}{5} \ }$ is not a solution.

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x - 8) = -2x – 4

-3x + 8 = -2x – 4

2x – 3x = -8 – 4

-x = -12

x = 12

Since $\boxed{ \ 12 \nless \frac{8}{3} \ }$ , $\boxed{ \ x = 12 \ }$ is not a solution.

Hence, the equation has no solution.

————————

<u>Problem (d)</u>

5|2x - 3| = 2|3 - 5x|

Let’s take the square of both sides. Then,

[5(2x - 3)]² = [2(3 - 5x)]²

(10x – 15)² = (6 – 10x)²

(10x - 15)² - (6 - 10x)² = 0

According to this formula $\boxed{ \ a^2 - b^2 = (a + b)(a - b) \ }$

$[(10x - 15) + (6 - 10x)][(10x - 15) - (6 - 10x)]] = 0$

(-9)(20x - 21) = 0

Dividing both sides by -9.

20x - 21 = 0

20x = 21

$x = \frac{21}{20}$

The only solution is $\boxed{ \ \frac{21}{20} \ }$

————————

<u>Problem (e)</u>

2x + |8 - 3x| = |x - 4|

We need to separate into four cases since we don’t know whether 8 – 3x and x – 4 are positive or negative. We cannot square both sides because there is a function of 2x.

8 – 3x is positive (or 8 - 3x > 0)
x – 4 is positive (or x - 4 > 0)

2x + 8 – 3x = x – 4

8 – x = x – 4

-2x = -12

x = 6

Substitute x = 6 into 8 – 3x ⇒ 8 – 3(6) < 0, it doesn’t work, even though when we substitute x = 6 into x - 4 it does work.

8 – 3x is positive (or 8 - 3x > 0)
x – 4 is negative (or x - 4 < 0)

2x + 8 – 3x = -(x – 4)

8 – x = -x + 4

x – x = = 4 - 8

It cannot be determined.

8 – 3x is negative (or 8 - 3x < 0)
x – 4 is positive. (or x - 4 > 0)

2x + (-(8 – 3x)) = x – 4

2x – 8 + 3x = x - 4

5x – x = 8 – 4

4x = 4

x = 1

Substitute x = 1 into 8 - 3x, $\boxed{ \ 8 - 3(1) \nless 0 \ }$ , it doesn’t work. Likewise, when we substitute x = 1 into x – 4, $\boxed{ \ 1 - 4 \not> 0 \ }$

8 – 3x is negative (or 8 - 3x < 0)
x – 4 is negative (or x - 4 < 0)

2x + (-(8 – 3x)) = -(x – 4)

2x – 8 + 3x = -x + 4

5x + x = 8 – 4

6x = 4

$\boxed{ \ x=\frac{4}{6} \rightarrow x = \frac{2}{3} \ }$

Substitute $x = \frac{2}{3} \ into \ 8-3x$ , $\boxed{ \ 8 - 3 \bigg(\frac{2}{3}\bigg) \not< 0 \ }$ , it doesn’t work. Even though when we substitute $x = \frac{2}{3} \ into \ x-4$ , $\boxed{ \ \bigg(\frac{2}{3}\bigg) - 4 < 0 \ }$ it does work.

Hence, the equation has no solution.

<h3>Learn more</h3>

The inverse of a function brainly.com/question/3225044
The piecewise-defined functions brainly.com/question/9590016
The composite function brainly.com/question/1691598

Keywords: hitunglah nilai x, the equation, absolute value of the function, has no solution, case, the only solution

5 0

3 years ago

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