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Alexxandr [17]
3 years ago
15

Write as one inequality with an absolute value.

Mathematics
1 answer:
Serjik [45]3 years ago
6 0
3|x-2|+1 = 16 is an inequality with an absolute value.
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What is the value of v? 45(v−7)=2
luda_lava [24]
<span>45(v−7)=2
45v - 315 = 2
45v = 317
v = 317/45
v = 7 2/45 (7 and 2/45)</span>
7 0
3 years ago
Read 2 more answers
Help me to do this please
Tanzania [10]

If x represents the mass of 1 ball in kg, then the balance shows ...

... 6x = x + 9 kg

... 5x = 9 kg

Then 6x is ...

... (6/5)·5x = (6/5)·9 kg = 10.8 kg

0.8 kg = 0.8·1000 g = 800 g

so 10.8 kg is ...

... D. 10 kg 800 g

7 0
3 years ago
What number is 10^2 times as long as 0.012?
Strike441 [17]
100, 3/250 because 10x10 is 100 and 0.012 converted into fraction is 3/250
7 0
3 years ago
A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.
kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is  x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is  −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is  y=3x−13  

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is  13 , the slope of a line perpendicular to line BC is  −3 . The equation of altitude 3 is  y=−3x−1  

The orthocenter is the point where all three altitudes intersect.

x=2  

y=3x−13  

y=−3x−1  

Use substitution to solve the first two equations  y=3(2)−13=−7  

The orthocenter is the point  (2,−7)  

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1  

−7=−6−1  

−7=−7  it works.

3 0
3 years ago
Question in screen shot
alisha [4.7K]
The two graphs in D are not inverses of each other.
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