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olganol [36]
3 years ago
8

what are the coordinates of the circumcenter of the triangle with vertices A(2,7), B(10,7) and C(10,3)

Mathematics
1 answer:
malfutka [58]3 years ago
5 0
Isusunfbfnfhdbdb dndbdb
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What can be concluded if <1≈ <7?
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B

Step-by-step explanation:

because the angles 1 and 7 are alternate and equal , we can conclude that p and q are parralel

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Write the given polynomial in standard form and write its degree .
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logan trades 50 pennies for nickels he gets a nickel for each group of 5 pennies what equation can be used to find how many nick
V125BC [204]

The  equation  that can be used to find how many nickels logan uses is: 50÷5.

<h3>Equation to find the number of nickels</h3>

Given:

Total pennies=50

Logan got a nickel of 5 pennies=1

Hence:

Number of nickels=1÷5×50

Number of nickels=50÷5

Number of nickels=10 nickels

Therefore the  equation  that can be used to find how many nickels logan uses is: 50÷5.

Learn more about Equation to find the number of nickels here:brainly.com/question/10704332

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4 0
2 years ago
Which trigonometric functions always have values less than 1? Explain your answer.
raketka [301]

Answer:

Notice that the values of sine and cosine are between 0 and 1. You found them by dividing the length of a leg by the hypotenuse. The hypotenuse is the longest side, so the numerator is less than the denominator. That means the output of the sine or cosine function is always less than 1.

Step-by-step explanation:

3 0
3 years ago
At which points are the tangents drawn to the ellipse x 2 + y 2 = [ a ] x + [ a ] y parallel to
In-s [12.5K]

The given equation of the ellipse is x^2 + y^2 = 2 x + 2 y

At tangent line, the point is horizontal with the x-axis therefore slope = dy / dx = 0

<span>So we have to take the 1st derivative of the equation then equate dy / dx to zero.</span>

x^2 + y^2 = 2 x + 2 y

x^2 – 2 x = 2 y – y^2

(2x – 2) dx = (2 – 2y) dy

(2x – 2) / (2 – 2y) = 0

2x – 2 = 0

x = 1

 

To find for y, we go back to the original equation then substitute the value of x.

x^2 + y^2 = 2 x + 2 y

1^2 + y^2 = 2 * 1 + 2 y

y^2 – 2y + 1 – 2 = 0

y^2 – 2y – 1 = 0

Finding the roots using the quadratic formula:

y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1

y = 1 ± 2.828

y = -1.828 , 3.828

 

<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828) and (1, 3.828).</span>

3 0
3 years ago
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