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zzz [600]
3 years ago
14

Explain how to use absolute value to find the distance between (-2,-2) and (-2,5)

Mathematics
1 answer:
dem82 [27]3 years ago
6 0
The absolute value of a number is its distance away from zero. Depending on how far away a number is, it has a less or greater absolute value. (-2,-2) have the same absolute value while (-2,5) have a different absolute value. Which has a greater absolute value? -25 or 6? The answer: -25, it is farther away from zero.
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Which best describes the graph of the cubic function f(x) = x3 + x2 + x + 1?
EleoNora [17]

Answer:

Option A

Step-by-step explanation:

  • f(x) = x³ + x² + x + 1

This is increasing function

The first option is correct

<em>See attached graph for reference</em>

3 0
3 years ago
What is the value of –21/6-1 1/4+1 3/4
aleksley [76]
It’s the answer pahaha
7 0
3 years ago
Read 2 more answers
12. What is the area of this figure?<br> 6 cm<br> 18<br> (57<br> 4 cm<br> 4 cm<br> 15 cm<br> 5 cm
adoni [48]

The area of this figure is 30 square cm.

Explanation

The figure consists of a rectangle and right angled triangle.

length of the rectangle l=6cm

breadth of the rectangle b=4 cm

area of the rectangle=lb=6*4 =24cm^2

hypotenuse of the right angled triangle=5 cm

height h=4 cm

we have to calculate the base using pythagoras theorem

b=\sqrt{H^2-h^2} =\sqrt{5^2-4^2} =\sqrt{25-16} =\sqrt{9} =3 cm

area of the triangle=1/2 bh\\=1/2 *3*4=6 cm^2

are of the figure=area of the rectangle+area of the triangle=24+6=30 sq cm

3 0
3 years ago
Approximate 17.850 litres to the nearest 1/2 litre
Alborosie
Approximating 17.850 litres with a precision to 1/2 (or 0.5) litre, you look at the number 8, and see if it lower or higher than 5. Since it is higher, you round it up, meaning you have 18 litres.

7 0
3 years ago
1.) How is the graph of y equals negative 2 x squared minus 5 different from the graph of y equals negative 2 x squared?
romanna [79]
-2x^2 - 5  is same shape as -2x^2 except that the whole graph is translated verticall downwards 5 units
5 0
3 years ago
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