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Bingel [31]
1 year ago
13

Suppose a triangle has sides a, b, and c, and that a2 + b2 < c2. Let be the measure of the angle opposite the side of length

c. Which of the following must be true? Check all that apply.
A. the triangle is not a right triangle.
B. a^2+b^2-c^2=2abcos(theta)
C. cos(theta) > 0
D. cos(theta) < 0
Mathematics
1 answer:
Sav [38]1 year ago
5 0

Answer:

the answer is ?

Step-by-step explanation:

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The correct answer would be, Jeremy rides at a greater speed than Kevin.

Step-by-step explanation:

Jeremy rides at a rate of 15 miles per hour

Kevin rides at a rate given in the table

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And X be the number of hours

Then for Jeremy:

y/x = 15/1

=> y= 15 x

For Kevin:  

(46-23)/(4-2)

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Complete the steps for solving 7 = –2x2 + 10x. Factor out of the variable terms. inside the parentheses and on the left side of
Mamont248 [21]

we have

7=-2x^{2} +10x

Factor the leading coefficient

7=-2(x^{2} -5x)

Complete the square. Remember to balance the equation by adding the same constants to each side

7-12.50=-2(x^{2} -5x+2.5^{2})

-5.50=-2(x^{2} -5x+2.5^{2})

Divide both sides by -2

2.75=(x^{2} -5x+2.5^{2})

Rewrite as perfect squares

2.75=(x-2.5)^{2}

Taking the square roots of both sides (square root property of equality)

x-2.5=(+/-)\sqrt{2.75}

Remember that

\sqrt{2.75}=\sqrt{\frac{11}{4}}= \frac{\sqrt{11}}{2}

x-2.5=(+/-)\frac{\sqrt{11}}{2}

x=2.5(+/-)\frac{\sqrt{11}}{2}

x=2.5+\frac{\sqrt{11}}{2}=\frac{5+\sqrt{11}}{2}

x=2.5-\frac{\sqrt{11}}{2}=\frac{5-\sqrt{11}}{2}

<u>the answer is</u>

The solutions are

x=\frac{5+\sqrt{11}}{2}

x=\frac{5-\sqrt{11}}{2}


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