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pishuonlain [190]
3 years ago
7

The area of a rectangular classroom is given by the trinomial a - 4a - 21. The length of the rectangle is a+3. What is the expre

ssion for the width of the classroom?
Mathematics
1 answer:
djverab [1.8K]3 years ago
3 0
I am assuming you wrote the problem wrong. Is it a^2?
A^2 -4A-21=0
(A-7)(A+3)=0
A=7 A=-3, ignore the negative number because it doesn't make sense if you plug in for a.
length=a+3=7+3=10
Width=7=a
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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 ship at position (1, 0) on a nautical chart (with north in the positive y direction) sights a rock at position (6, 5). What is
OLga [1]

Answer:

The vector joining the ship to the rock is t= 7 i + 5 j

The direction is 0.9505 radians east of north.

Step-by-step explanation:

The position vector of the ship:

r= 1 i + 0 j

The position vector of the ship:

s= 6 i + 5 j

The vector joining the ship to the rock is:

t = r + s

t = (1 i + 0 j) + (6 i + 5 j)

t = 7 i + 5 j

The bearing of the rock to the ship is:

Θ= \frac{ \pi}{2} - arctan (\frac{5}{7})= 0.9505 radians

4 0
3 years ago
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Step-by-step explanation:

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