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3 years ago

| x | = 6 Solve for x

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

1 x 6= 6 x 1= 6

Ronch [10]3 years ago

3 0

Answer:

x = 6, -6

Step-by-step explanation:

The absolute value of a number is its distance from 0 on the number line. To find the absolute value of a number through an equation, write two equations: one where the side without the variable is positive, and one where the side without the variable is negative. In this case, it would look like this:

| x | = 6 and | x | = -6

Then, solve whatever is inside the absolute value and take out the absolute value lines. In this equation, we don't need to solve anything, so we can just take out the lines. So, we'll end up with:

x = 6 and x = -6

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How do I solve this systems of three equations problem?

Serhud [2]

Answer:

You have to eliminate one of the variables when the equations are added.

A) x + 2y + z = 10

B) 2x -y +3z = -5

C) 2x -3y -5z = 27

We mulutiply A) by 2 then add it to B and C

A) 2x + 4y + 2z = 20 and the sum =

6x = 42 Luckily, the "y" and "z" variables cancel out and we find:

x = 7

Then we use x =7 in calcucating equation A) and B)

A) 2y + z = 3

B) -y + 3z = -19

THEN solve for y and z by eliminating variables.

Step-by-step explanation:

3 0

3 years ago

The mean cost of a five pound bag of shrimp is 46 dollars with a variance of 64. If a sample of 53 bags of shrimp is randomly se

sertanlavr [38]

Answer:

The probability that sample mean differ the true greater than 2.1 will be 2.8070 %

Step-by-step explanation:

Given:

Sample mean =46 dollars

standard deviation=8

n=53

To Find :

Probability that sample mean would differ from true mean by greater than 2.1

Solution;

This sample distribution mean problem,

so for that

calculate Z- value

Z=(sample mean - true mean)/(standard deviation/Sqrt(n))

Z=-2.1/(8/Sqrt(53))

Z=-2.1*Sqrt(53)/8

Z=-1.91102

Now for P(X≥2.1)=P(Z≥-1.91102)

Using Z-table,

For Z=-1.91

P(X>2.1)=0.02807

6 0

3 years ago

Perform the indicated operations. Write the answer in standard form, a+bi. 5-3i / -2-9i

Vsevolod [243]

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i} \\$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)}) \\$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$ .

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}}) \\$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85}) \\$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85}) \\$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$ .

$\large \bf \: Re(\frac{-10+45i+6i+27}{85}) \\$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85}) \\$

Do the additions in $\sf-10+27+\left(45+6\right)i$ .

$\large \bf Re(\frac{17+51i}{85}) \\$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i \\$ .

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i) \\$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i \\$ is $\sf \frac{1}{5} \\$ .

$\large \boxed{\bf\frac{1}{5} = 0.2} \\$

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3 years ago

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3 years ago

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yawa3891 [41]

Answer:

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