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AURORKA [14]
3 years ago
9

Help ASAP I don’t get it

Mathematics
1 answer:
GarryVolchara [31]3 years ago
3 0

9514 1404 393

Answer:

  • positive
  • negative

Step-by-step explanation:

The thrust of the question is to make sure you understand that increasing the y-coordinate of a point will move the point upward, and decreasing it will move the point downward.

That is adding a positive value "k" to x^2 will move the point (x, x^2) to the point (x, x^2+k), which will be above the previous point by k units.

If k is subtracted, instead of added, then the point will be moved downward.

The blanks are supposed to be filled with <u> positive </u>, and <u> negative </u>.

_____

<em>Comment on the question</em>

The wording of the statement you're completing is a bit odd. If k is negative (-2, for example), this statement is saying the graph is translated down -2 units. It is not. It is translated down |-2| = 2 units. The direction of translation depends on the sign of k. The amount of translation depends on the magnitude of k.

If you thoroughly understand (x, y) coordinates and how they are plotted on a graph, it should be no mystery that changing the y-coordinate will change the vertical position of the graph.

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Can u help me I don't get it
Elina [12.6K]
Equivalent fractions can you help add or subtract because it can be easier to simplify.


hope that helped
6 0
3 years ago
Fred and Ben each ran laps around the track. Fred completed his run in 16 minutes. Ben completed his run in 3/4 of the amount of
m_a_m_a [10]

Answer:

Time required by Ben was 12 minutes.

Step-by-step explanation:

Given:

Time Required for Fred = 16 mins

Ben completed his run in 3/4 of the amount of time of Freds run.

We need to find the time required by Ben.

Now Given that;

Ben completed his run in 3/4 of the amount of time of Freds run.

It means that time required by Ben is equal to 3/4  times time required by Fred.

Time required by Ben = \frac{3}{4}\times  Freds \ run

Time required by Ben = \frac{3}{4}\times 16 = 12\ mins

Hence Time required by Ben was 12 minutes.

7 0
3 years ago
How do you do this?<br> Please help, I’ve been staring at this for like 10 min :)
damaskus [11]

Answer & Step-by-step explanation:

The formula is given in the middle. You are supposed to insert the value of x into this formula to find the value of y (like in the first given example):

x=-2\\y=2(-2)+5\\y=-4+5\\y=1

------

x=-1\\y=2(-1)+5\\y=-2+5\\y=3

------

x=0\\y=2(0)+5\\y=0+5\\y=5

------

x=1\\y=2(1)+5\\y=2+5\\y=7

------

x=2\\y=2(2)+5\\y=4+5\\y=9

:Done

3 0
4 years ago
Approximently how many inches are in 2500 millimeters
GarryVolchara [31]
98.4252<span> inches is the exact so about 98 inches</span>
4 0
3 years ago
Read 2 more answers
At a recent marathon, spectators lined the street near the starting line to cheer for the runners. The crowd lined up 5 feet dee
masha68 [24]

Answer:

29568 people cheered for the runners at the start of the race

Step-by-step explanation:

From the question, the crowd lined up 5 feet deep on both sides of the street for the first mile.

This lined up crowd could be related to a rectangle that is 1 mile long and 5 feet wide.

First, Convert 1 mile to feet

1 mile = 5280 feet

Hence, the length of the rectangle is 5280 feet and the width is 5 feet.

Now, we will determine how many 5 feet by 5 feet square we can get from the 5280 feet by 5 feet rectangle. To do that, we will divide 5280 feet by 5 feet

5280 feet ÷ 5 feet = 1056

Hence, from the rectangle, we can get 1056 5 feet by 5 feet square.

From the question, you estimate that 14 people can comfortably fit in a square that measures 5 feet by 5 feet,

∴ 14 × 1056 people will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

14 × 1056 = 14784 people

This is the amount of people that will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

Since the crowd lined up on both sides of the street, then

2 × 14784 people will comfortably fit in the crowed lined up 5 feet deep on both sides of the street for the first mile

2 × 14784 = 29568 people

Hence, 29568 people cheered for the runners at the start of the race.

5 0
4 years ago
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