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3 years ago

QUESTION 16 What is the absolute value of -72?

Mathematics

2 answers:

Serjik [45]3 years ago

7 0

Answer:

Step-by-step explanation:

l-72l is equal to 72

The absolute value of any number is the positive of that number.

hope it helps :).

i need brainliest to level up *hint* *hint*

11Alexandr11 [23.1K]3 years ago

3 0

The absolute value of -72 is 72.

Absolute value is essentially just making the number given positive.

If the given number is already positive then the absolute value is the same as the original number

Hope this helped!

~Just a girl in love with Shawn Mendes

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Cho ma trận A và một sự tương ứng

algol [13]

Answer:

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2 years ago

a right triangle is removed from a rectangle to create the shaded region shown below. find the area of the shaded region. be sur

ser-zykov [4K]

Answer:

Step-by-step explanation:

Objective: Find missing dimensions of the right triangle using the then subtract that area of the triangle from the rectangle's area.

Using the definition of a rectangle, the left vertical side measure is 7 and right vertical side is 6. So the missing legs of the triangle is 4 and 3.

Apply triangle formula, base x height divided by 2.

$\frac{4 \times 3}{2} = 6$

Area of rectangle is length x width so the area is

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Subtract 6 from 42.

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2 years ago

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horsena [70]

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6 0

3 years ago

Y=+/-√bx+c+d Use the inverse of the function y= x2 -18x to find the unknown values.<br>

ozzi

Answer:

a=1

b=81

c=9

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

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3 years ago