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3 years ago

Can an absolute value equal a negative?

1 answer:

Montano1993 [528]3 years ago

7 0

The distance of it away from zero cannot be a negative distance, distance is always positive therefor the absolute value will always be positive.

Short answer: no, absolute value is always positive

I hope this helps :)

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Find the area of this,<br> Simplify your answer completely.

creativ13 [48]

Answer:

$\frac{2}{15} {cm}^{2}$

Step-by-step explanation:

Area of Rectangle = Length x Breadth

$\frac{1}{5} \times \frac{2}{3} \\ = \frac{2}{15} {cm}^{2}$

4 0

2 years ago

Help? I don't understand how to do this!

igomit [66]

87.96 You need the height of the circle which is
14 because 7 is only the radius. And you put it into the equation and solve C=2πr

4 0

3 years ago

Restrict the domain of f(x) = |x − 3|2 + 2 on which the function is invertible.

andrew-mc [135]

The function is invertible on either side of the vertex. There are many possible answers. Ones that include the vertex are ...

x ≥ 3
or
x ≤ 3

8 0

3 years ago

What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?

kompoz [17]

Answer:

K(-4,8) is the ortho center.

Step-by-step explanation:

In a right angled triangle, The vertex of the right angle is the ortho center.

Here we are given

J(-4,-1), K(-4,8) & L(2,8)

Using distance formula we get

$JK^{2}= (-4+4)^{2}+ (8+1)^{2}=0+81=81$

$KL^{2}= (-4-2)^{2}+ (8+-8)^{2}=36+0=36$

$JL^{2}= (-4-2)^{2}+ (8+1)^{2}=36+81=117$

So we can say that

$JK^2 + KL^2=JL^2$

By converse of pythagorean theorem we get

Hence the Vertex of the right angle is K(-4,8)

K(-4,8) is the ortho center.

7 0

3 years ago

Only question 10. please

Klio2033 [76]

For a, the circle would be quadrupled and the circumference will also be doubled

For b, the circumference will also be tripled

6 0

3 years ago