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notsponge [240]
3 years ago
5

Find the coordinates of the circumcenter of the triangle below

Mathematics
1 answer:
Lena [83]3 years ago
6 0
(2,2) (6,2) (6,8) Hope this helps I’m pretty sure this is correct :)
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Determine the factors of x2 − 12x − 20.
FrozenT [24]

Answer:

Step-by-step explanation:

Synthetic division is one way to determine whether or not a given number is a root of the quadratic.  x^2 − 12x − 20 can be rewritten as x^2 - 12x + 36 - 36 - 20, or (x - 6)^2 - 56, which does not have integer solutions:

(x - 6)^2 - 56 = 0 becomes (x - 6)^2 = 56, which works out to x - 6 = ± 2√14.

None of the possible roots suggested in this problem turns out to be an actual root.

correct response: PRIME

3 0
3 years ago
4 ALGEBRA 1B QUESTIONS - 70PTS
max2010maxim [7]
They have two solutions
6 0
3 years ago
How much money does Conor earn each minute he does yard work, if he works 3.25 hours at $9 an hour?
LenaWriter [7]

Answer:

$2.76

Step-by-step explanation:

9/3.25=2.76

8 0
2 years ago
A well mixed tank with a capacity of 1500 gals originally contains 1000 gallons of freshwater. One pipe containing 1/2 lb of sal
Illusion [34]

Answer:

400 lb of salt

Step-by-step explanation:

Let us assume the water flows into the rank for x minutes.

There is an initial of 1000 gallons of water in the tank and water flows in through one pipe at 4 gal/min and through another pipe at 6 gal/min. In x minute, the amount of water in the tank = 1000 + 4x + 6x = 1000 + 10x

Water flows out at 5 gal/min, therefore in x minute the amount of water in the tank = 1000 + 10x - 5x = 1000 + 5x

The tank begins to overflow when it is full (has reached 1500 gallons). Therefore:

1500 = 1000 + 5x

5x = 1500 - 1000

5x = 500

x = 100 minutes.

1/2 lb salt per gallon flows into the tank at 4 gal/min and 1/3 lb of salt is flowing in at 6 gal/min, in 100 min the amount of salt that entered the tank = 4 gal/min × 100 min × 1/2 lb/gal + 6 gal/min × 100 min × 1/3 lb/gal= 400 lb

Therefore the amount of salt is in the tank when it is about to overflow = 400 lb of salt

7 0
3 years ago
Two cables support a 800​-lb ​weight, as shown. Find the tension in each cable.
jeka94

Answer:

  • 892 lb (right)
  • 653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

  Rcos(45°) +Lcos(75°) = 800

  Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

  R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

     = 800sin(75°)/sin(120°) ≈ 892 . . . pounds

  L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

_____

This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

  (right, left) = w/sin(α+β)×(sin(β), sin(α))

5 0
2 years ago
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