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irakobra [83]
3 years ago
6

What is the equivilent of +367 on a numberline?

Mathematics
1 answer:
Mama L [17]3 years ago
8 0

Answer:

367

Step-by-step explanation:

367 because it is positive

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Y = x + 5<br> 2x + y = 17
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Answer:

y = x + 5: x=y−5

2x + y = 17 : x=-1/2y+17/2

Step-by-step explanation:

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The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) us
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horrorfan [7]

Answer:

is that dababy?

Step-by-step explanation:

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2 years ago
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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

||a|| = √130

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
3 years ago
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