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NeTakaya
3 years ago
8

Write the coordinates of the point C after dilation with scale factor of 5, centered at the orgin

Mathematics
1 answer:
Ber [7]3 years ago
7 0

Answer:

It's C buddy

Step-by-step explanation:

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1. Which relation is a function?
Amiraneli [1.4K]

Answer:

1. (0, 4), (-4, 2), (7, 1), (-8, 2)

2. (2, 4), (-3, 2), (9, 1), (-7, 2)

Step-by-step explanation:

1. Since there is 1 value of y for every value of x in (0, 4), (-4, 2), (7, 1), (-8, 2) this relation is a function.

2. Since there is 1 value of y for every value of x in (2, 4), (-3, 2), (9, 1), (-7, 2) this relation is a function.

5 0
3 years ago
Read 2 more answers
What is the sum of 5/30+4/6
vovangra [49]
5/30+4/6
5/30=1/6
1/6+4/6=5/6
5/6
Hope I helped!
6 0
3 years ago
Stem and leaf plots, what is the mode median and range of the scores ​
frez [133]

Answer:

Mode = 81,  Median = 81 and Range = 39.

Step-by-step explanation:

The mode  is the most occurring value which is 81.

The median is also 81. There are a total of 19 numbers arranged in ascending order so the median is the 10th number.

The range is the highest - lowest number = 100 - 61 = 39.

8 0
4 years ago
the school cafeteria sells two kinds of wraps: vegetarian and chicken. the vegetarian wraps costs $1.00 and the chicken wrap cos
artcher [175]
Help on my most recent question
3 0
3 years ago
Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i
Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333

So the correct answer is C, the area between the two functions is 4/3 units.

3 0
3 years ago
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