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wolverine [178]
3 years ago
14

Find the distance a(3, -1) b(0, -4)

Mathematics
1 answer:
MrRissso [65]3 years ago
5 0

Answer:

Distance =

            \sqrt{(3-0)^2+(-1-(-4))^2}  = \sqrt{9 +(-1+4)^2} =\sqrt{9+9}=\sqrt{18}=3\sqrt{2}

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Find the coordinates of the circumcenter for triangle DEF with coordinates D(1,3) E(8,3) and F(1,-5) show your work
nikklg [1K]

One leg is horizontal, on the line y=3. Another leg is vertical, on the line x=1. The point where these intersect, D(1, 3) is the vertex of a right angle. Any right triangle inscribed in a circle has its hypotenuse as the diameter of the circle.

Since the triangle is a right triangle, the circumcenter is the midpoint of the hypotenuse: ((8, 3) + (1, -5))/2 = (4.5, -1).

6 0
3 years ago
Please help step by step!!!
Lesechka [4]

Answer:

-7

Step-by-step explanation:

We first are going to flip the equation around.

So instead of 9 times x = -36 we'll do 36/9. 36/9 = -7.

Also you will kind of know it will be a negative number from the beginning since it's -36.

I don't know if that last part makes any sense, but I hope this helps you!

5 0
2 years ago
Read 2 more answers
Which choices are real numbers?
Snezhnost [94]
 a and c are your answers
8 0
3 years ago
A bag contains 6 RED beads, 3 BLUE beads, and 11 GREEN beads. If a single bead is picked at random, what is the probability that
Anna11 [10]

Answer:

17/20

Step-by-step explanation:

Total no. of beads = 6 + 3 + 11 = 20

We need RED or GREEN bead .

So no. of beads needed = 6 + 11 = 17

So probability of getting GREEN or RED beads = 17/20

6 0
3 years ago
Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab
Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

<u>Left part:</u>

\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

<u>Right part:</u>

\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

Since simplified left and right parts are the same, then the equality is true.

3 0
3 years ago
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