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Natalija [7]
4 years ago
9

This table gives a few (x,y) pairs of a line in the coordinate plane. What is the y-intercept of the line? (full problem attache

d)

Mathematics
2 answers:
velikii [3]4 years ago
7 0

Answer:

(0,34)

Step-by-step explanation:

For each rise of 14 in the x direction, this graph rises by -8 in the y direction. This means that, when x is 0, and the graph intersects the y axis, the y value will be 50-8-8=34. Therefore, the y intercept of this line is (0,34). Hope this helps!

Bezzdna [24]4 years ago
6 0

Answer:

The answer is (0,34)

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Which fraction is equivalent to 16/20 ?A.8/100 <br> B.6/10<br> C.48/60<br> D.20/24
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Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k​
valkas [14]

Answer:

\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k​, the vector projection of B unto a is expressed as proj_ab = \dfrac{b.a}{||a||^2} * a

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k  =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

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square both sides

||a||² = (√130)

||a||²  = 130

proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\

<em>Hence the projection of b unto a is expressed as </em>\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\<em></em>

7 0
4 years ago
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