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Brut [27]
3 years ago
11

What is the midpoint of the line segment with endpoints at (12,7)and (18,19)

Mathematics
1 answer:
Mrac [35]3 years ago
6 0

Answer:

(15, 13)

Step-by-step explanation:

Midpoint is the half of the segment.

What I did to solve this was adding the two xs together to get 30 then find the average of 30 by dividing it by 2. This would give you 15

I did the same thing with the ys and got 13

hope this made sense

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What is the solution to he equation (y/y-4)-(4/y+4)=3^2/y^2-16
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Answer:

y=\pm i \sqrt{7}

Step-by-step explanation:

Given equation is \frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}

Factor denominators then solve by making denominators equal

\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}

\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}

\frac{y\left(y+4\right)-4\left(y-4\right)}{\left(y-4\right)\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}

y\left(y+4\right)-4\left(y-4\right)=9

y^2+4y-4y+16=9

y^2=9-16

y^2=-7

take squar root of both sides

y=\pm \sqrt{-7}

y=\pm i \sqrt{7}

Hence final answer is y=\pm i \sqrt{7}.

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Which of the following polynomials is in standard form?
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Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 5.08 and angle θA = 25.6 ∘ measured in the sense from the +x
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Answer:

(A⃗ ×B⃗ )⋅C⃗ = - 76.415

Step-by-step explanation:

First we need to calculate (A⃗ ×B⃗ ) :

(A⃗ ×B⃗ ) = A.B.sin (α).n

Where A is the magnitude of A⃗

Where B is the magnitude of B⃗

Where α is the angle between A⃗ and B⃗ = 63.9 - 25.6 = 38.3

Finally n is the vector orthogonal to A⃗ and B⃗

n magnitude is 1 and his direction is given by the right hand-rule

so n = ( 0 , 0 , 1 )

(A⃗ ×B⃗ ) = A.B.sin (α).n = 5.08 . 3.94 . sin (38.3) . (0 , 0 , 1 ) = (0,0,12.4)

C⃗ can be written as C.(0,0,-1) because of his +z - direction

C.(0,0,-1) = 6.16.(0,0,-1) = (0,0,-6.16)

(A⃗ ×B⃗ )⋅C⃗ = (0,0,12.4).(0,0,-6.16) = -76.41480787 = -76.415

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