3 years ago

Solve for x: 2x^2 + x - 4 = 0

Mathematics

1 answer:

ella [17]3 years ago

7 0

Answer:

Hopefully this helps, sorry if it doesn't.

Step-by-step explanation:

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Which of the following is always true about Vertical Angles?

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3 years ago

Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

3 years ago

Read 2 more answers

Let f(x) = x + 8 and g(x) = x2 − 6x − 7. find f(g(2)). −7 −3 10 33 open study

Ray Of Light [21]

F(g(2))

g(x) = x^2 - 6x - 7
g(2) = 2^2 - 6(2) - 7
g(2) = 4 - 12 - 7
g(2) = -15

f(x) = x + 8
f(g(2) = f(-15) = -15 + 8 = -7 <==

ur answer is -7

6 0

3 years ago