Find the distance between two points (-6,2) and (2,7)

Mathematics

1 answer:

Sever21 [200]3 years ago

7 0

Answer:

√89 (decimal: 9.43398...)

Step-by-step explanation:

Call one point Point 1 (x1,y1) and make the other Point 2 (x2,y2).

apply rule -(-a) = a

√(2 + 6)^2 + (7 - 2)^2

add the numbers: 2 + 6 = 8

√8^2 + (7 - 2)^2

subtract the numbers: 7 - 2 = 5

√8^2 + 5^2

8^2 = 64

√64 + 5^2

5^2 = 25

√64 + 25

add the numbers: 64 + 25 = 89

= √89

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Use the discriminant to determine how many real number solutions exist for the quadratic equation –4j2 + 3j – 28 = 0. A. 3 B. 2

Zepler [3.9K]

Answer:

C. 0

Step-by-step explanation:

–4j^2 + 3j – 28 = 0

The discriminant is b^2-4ac if >0 we have 2 real solutions

=0 we have 1 real solutions

<0 we have 2 imaginary solutions

a = -4, b =3 c = -28

b^2 -4ac

(3)^2 - 4(-4)*(-28)

9 - 16(28)

9 -448

This will be negative so we have two imaginary solutions.

Therefore we have 0 real solutions

5 0

3 years ago

Help me please

kari74 [83]

Answer:

$\frac{ \cos(80) }{ \sin(10) } + \frac{ \sin(20) }{ \cos(70) } = 2 \\ \frac{ \cos(90 - 10) }{ \sin(10) } + \frac{ \sin(20) }{ \cos(9 0 - 20) } = 2 \\ \frac{ \sin(10) }{ \sin(10) } + \frac{ \sin(20) }{ \sin(20) } = 2 \\ 1 + 1 = 2 \\ 2 = 2 \\ \\ \frac{ \cot(40) }{ \tan(50) } + \frac{ \cos(65) }{ \sin(115) } = 2 \\ \frac{ \cot(90 - 50) }{ \tan(50) } + \frac{ \cos(65) }{ \sin(90 + 65) } = 2 \\ \frac{ \tan(50) }{ \tan(50) } + \frac{ \cos(65) }{ \cos(65) } = 2 \\ 1 + 1 \\ = 2$