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3 years ago

What is the absolute value of –38? A. 19 B. –19 C. 38 D. –38

Mathematics

1 answer:

GREYUIT [131]3 years ago

6 0

<u>Answer</u>

C. 38

<u>Explanation</u>

The absolute value of any digit is the magnitude or its size. The negative side is not considered.

It can also be defined as the actual magnitude of a numerical value or measurement, irrespective of its relation to other values.

It is represented as; /x/.

The absolute value of -38 is /-38/.

∴ /-38/ = 38

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Read 2 more answers

Need help with homework

motikmotik

We have two points describing the diameter of a circumference, these are:

$\begin{gathered} A=(-12,-4) \\ B=(-4,-10) \end{gathered}$

Recall that the equation for the standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where (h,k) is the coordinate of the center of the circle, to find this coordinate, we find the midpoint of the diameter, that is, the midpoint between points A and B.

For this we use the following equation:

$M=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})$

Now, we replace and solve:

$\begin{gathered} M=(\frac{-12+(-4)}{2},\frac{-4+(-10)}{2} \\ M=(\frac{-12-4}{2},\frac{-4-10}{2}) \\ M=(\frac{-16}{2},\frac{-14}{2}) \\ M=(-8,-7) \end{gathered}$

The center of the circle is (-8,-7), so:

$\begin{gathered} h=-8 \\ k=-7 \end{gathered}$

On the other hand, we must find the radius of the circle, remember that the radius of a circle goes from the center of the circumference to a point on its arc, for this we use the following equation:

$r^2=\Delta x^2+\Delta y^2$

In this case, we will solve the delta with the center coordinate and the B coordinate.

$\begin{gathered} r^2=((-4)-(-8))^2+((-10)-(-7)) \\ r^2=(-4+8)^2+(-10+7)^2 \\ r^2=4^2+(-3)^2 \\ r^2=16+9 \\ r^2=25 \\ r=5 \end{gathered}$

Therefore, the equation for the standard form of a circle is:

$\begin{gathered} (x-(-8))^2+(y-(-7))^2=25 \\ (x+8)^2+(y+7)^2=25 \end{gathered}$

In conclusion, the equation is the following:

$(x+8)^2+(y+7)^2=25$

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