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3 years ago

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 2x + y = 17

larisa [96]

Answer:

y = x + 5: x=y−5

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

Step-by-step explanation:

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3 years ago

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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I believe answer is 7,0/ hope this helps

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3 years ago

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What is 5+5 Someone plzzz help me Sike

horrorfan [7]

Answer:

is that dababy?

Step-by-step explanation:

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2 years ago

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3 years ago

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

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

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

7 0

3 years ago