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3 years ago

Which of the following should be considered in buying a computer?

Mathematics

2 answers:

koban [17]3 years ago

8 0

Answer:

All of the following should be considered when buying a computer.

ki77a [65]3 years ago

5 0

All of them is needed

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Show the inequality for x. Show each step of the solution 12x>7(4x+5)-19

Dima020 [189]

Answer:

x<-1

Step-by-step explanation:

12x>7(4x+5)-19

Solve the right side

12x > 28x+35-19

12x > 28x+16

Make x the subject

12x-28x > 16

-16x > 16

-x > 16/16

-x > 1

Multiplying by -1 on both sides

x < -1

If you multiply ">" with minus sign, its direction changes. It is similar to switching the positions of x and -1.

Simplify

x < -1

The inequality of x is x<-1

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3 years ago

A 2.0 kg pendulum swings from point A of height YA

Vedmedyk [2.9K]

Answer: =0.39 J

Step-by-step explanation:

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3 years ago

Y=(x-1)^2+2 find the y intercept

Elden [556K]

Is this slope-intercept form?

5 0

3 years ago

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The quadratic equation 8x²+12x-14 has two real roots. What is the sum of the squares of these roots?

mars1129 [50]

Answer:

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

The sum of the squares of these roots is $\frac{-3}{2}$

Step-by-step explanation:

The given quadratic equation is $8x^2+12x-14$ has two real roots.

To find the roots .

$8x^2+12x-14=0$

Dividing the above equation by 2

$\frac{1}{2}(8x^2+12x-14)=\frac{0}{2}$

$4x^2+6x-7=0$

For quadratic equation $ax^2+bx+c=0$ the solution is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a and b are coefficents of $x^2$ and x respectively, c is a constant.

For given quadratic equation

a=4, b=6, c=-7

$x=\frac{-6\pm\sqrt{6^2-4(4)(-7)}}{2(4)}$

$=\frac{-6\pm\sqrt{36+112}}{8}$

$=\frac{-6\pm\sqrt{148}}{8}$

$=\frac{-6\pm\sqrt{37\times 4}}{8}$

$=\frac{-6\pm\sqrt{37}\times\sqrt{4}}{8}$

$=\frac{-6\pm\sqrt{37}\times 2}{8}$

$=2\frac{(-3\pm\sqrt{37})}{8}$

$=\frac{-3\pm\sqrt{37}}{4}$

$x=\frac{(-3\pm\sqrt{37})}{4}$

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

Now to find the sum of the squares of these roots

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3+\sqrt{37}-3-\sqrt{37}}{4}$

$=\frac{-6}{4}$

$=\frac{-3}{2}$

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3}{2}$

Therefore the sum of the squares of these roots is $\frac{-3}{2}$

3 0

3 years ago

What is Q3 for this data set?

Kisachek [45]

The same as last time

Step-by-step explanation:

8 0

2 years ago

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