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3 years ago

1. Find the equation of a line that is perpendicular to y=-2/3+4 and goes through (0,5)

2 answers:

7 0

Y = 3/2 + 5
since it’s perpendicular, flip the numerator and denominator of the slope and b=5 because it’s going through the y-axis at (0,5)

nydimaria [60]3 years ago

6 0

Answer:

Step-by-step explanation:

It says perpendicular, which means if you graphed the line given and the one you have to make, they would intersect at a right angle. I'm going to presume the x is after -2/3 since you didn't put one in, I hope that's correct.

The first part to solving problems like these is to invert the slope, which just means you have to switch the sign on it (negative to positive/positive to negative) and switch the numerator and denominator. This makes the new slope 3/2.

The next step is to find the y-intercept, which they thankfully gave you in this problem (when x=0 the y-value is always the y intercept), but you'll usually have to find it.

The equation is y=3/2x+5. You should double check this, I'm good at math but it's been a bit.

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3 years ago

A child walks due east on the deck of a ship at 3 miles perhour.

VladimirAG [237]

Answer:

16.28 mph ; 79.38°

Step-by-step explanation:

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4 years ago

Find all unit vectors that are orthogonal to the vector u = 1, 0, −4 .

KIM [24]

Answer:

Step-by-step explanation:

Given:

u = 1, 0, -4

In unit vector notation,

u = i + 0j - 4k

Now, to get all unit vectors that are orthogonal to vector u, remember that two vectors are orthogonal if their dot product is zero.

If v = v₁ i + v₂ j + v₃ k is one of those vectors that are orthogonal to u, then

u. v = 0 [<em>substitute for the values of u and v</em>]

=> (i + 0j - 4k) . (v₁ i + v₂ j + v₃ k) = 0 [<em>simplify</em>]

=> v₁ + 0 - 4v₃ = 0

=> v₁ = 4v₃

Plug in the value of v₁ = 4v₃ into vector v as follows

v = 4v₃ i + v₂ j + v₃ k -------------(i)

Equation (i) is the generalized form of all vectors that will be orthogonal to vector u

Now,

Get the generalized unit vector by dividing the equation (i) by the magnitude of the generalized vector form. i.e

$\frac{v}{|v|}$

Where;

|v| = $\sqrt{(4v_3)^2 + (v_2)^2 + (v_3)^2}$

|v| = $\sqrt{17(v_3)^2 + (v_2)^2}$

$\frac{v}{|v|}$ = $\frac{4v_3i + v_2j + v_3k}{\sqrt{17(v_3)^2 + (v_2)^2}}$

This is the general form of all unit vectors that are orthogonal to vector u

where v₂ and v₃ are non-zero arbitrary real numbers.

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3 years ago