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sergiy2304 [10]
2 years ago
7

Are the following equations parallel, perpendicular, or neither?

Mathematics
1 answer:
m_a_m_a [10]2 years ago
7 0

Answer:

neither one of them that solves the problem

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If the coordinates of the endpoints of a diameter of the circle are​ known, the equation of a circle can be found.​ First, find
Evgesh-ka [11]

Answer:

The equation of the circle is (x+3)^2+(y-5)^2 = 17

Step-by-step explanation:

The complete question is

If the coordinates of the endpoints of a diameter of the circle are​ known, the equation of a circle can be found.​ First, find the midpoint of the​ diameter, which is the center of the circle. Then find the​ radius, which is the distance from the center to either endpoint of the diameter. Finally use the​ center-radius form to find the equation.

Find the center-radius  form of the circle having the points (1,4) and (-7,6) as the endpoints of a diameter.

Consider that, if both points are the endpoints of a diameter, the center of the circle is the point that is exactly in the middle of the two points (that is, the point whose distance to each point is equal). Given points (a,b) and (c,d), by using the distance formula, you can check that the middle point is the average of the coordinates. Hence, the center of the circle is given by

(\frac{1-7}{2}, \frac{4+6}{2}) = (-3,5).

We will find the radius. Recall that the radius of the circle is the distance from one point of the circle to the center. Recall that the distance between points (a,b) and (c,d) is given by \sqrt[]{(a-c)^2+(b-d)^2}. So, let us use (1,4) to calculate the radius.

r = \sqrt[]{(1-(-3))^2+(4-5)^2} = \sqrt[]{17}.

REcall that given a point (x_0,y_0). The equation of a circle centered at the point (x_0,y_0) is

(x-x_0)^2+(y-y_0)^2 = r^2

In our case, (x_0,y_0)=(-3,5) and r=\sqrt[]{17}. Then, the equation is

(x-(-3))^2+(y-5)^2 = (x+3)^2+(y-5)^2 = 17

4 0
3 years ago
A camp needs 100 students to help with 4-year-old campers. Eight students from 4 different classes have agreed to help. How many
dimaraw [331]

Answer: 32 students

Step-by-step explanation: 8 x 4 = 32  

4 0
3 years ago
Which of the following measures cannot be determined from a box plot?
ad-work [718]
<h3>3 Answers:</h3>
  • B) Mean
  • C) Mean absolute deviation
  • E) Mode

==========================================================  

Explanation:

The box plot, aka "box-and-whisker plot", visually represents five things. These things are:

  • Minimum
  • Q1 = first quartile
  • Median (sometimes referred to as Q2 or second quartile)
  • Q3 = third quartile
  • Maximum

This list of five items is known as the five number summary.

The min is the tip of the left most whisker, assuming there aren't any small outliers. The max is the opposite side, being the tip of the right most whisker (assuming no large outliers). If there are any outliers, then they'll be shown as "island" dots on their own separated from the main box plot. The left and right edges of the box are Q1 and Q3 respectively. The median is the vertical line inside the box. The vertical line does not have to be at the midpoint of the left and right edges of the box. It simply needs to be somewhere in the box.

--------

Since the box plot lets us know the min and max, we can compute the range because

range = max - min

and we can also calculate the interquartile range (IQR) because

IQR = Q3 - Q1

--------

So to summarize so far, the five number summary is visually represented as the box plot. The range and IQR can be computed using items from the five number summary.

We cannot compute the mean because we would need the actual data set of values, rather than the summary data. The same goes for the mean absolute deviation (MAD) and the mode. Since your teacher is looking for things that cannot be determined from a box plot, we'll go for answers B, C and E.

In other words, we rule out choices A, D, and F because we can compute or determine those values from a box plot.

7 0
3 years ago
Solving systems of equations<br><br> X-2y=0<br> Y=2x-3
bearhunter [10]
\left \{ {{x-2y=0} \atop {y=2x-3}} \right. \\\\Substitute\ second\ equation\ to\ first:\\\\x-2(2x-3)=0\\x-4x+6=0\\-3x=-6\ \ \ \ \ |:-3\\x=2\\\\y=2\cdot2-3\\y=4-3=1\\\\ \left \{ {{y=1} \atop {x=2}} \right.
5 0
3 years ago
Read 2 more answers
Hello can you please help me posted picture of question
zalisa [80]
Its B. 2/13, becouse you do the math correctly and you will find it eventually
6 0
3 years ago
Read 2 more answers
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