Draw a vector representing the course of a ship that goes 75 miles on a
1 answer:
Answer:
The answer is below
Step-by-step explanation:
We have a vector with a magnitude of 75 miles and a direction of 300 °.
To draw it, the first thing is to make a Cartesian plane, the first thing is to know in which quadrant it would be, 300 ° is in the fourth quadrant (270 ° - 360 °), that is, it would be 30 ° (300 ° - 270 °) with respect to to the y axis negative.
Knowing this angle, we calculate the value of x and y, knowing that the magnitude of the vector is 75 miles, that is, this value would be the hypotenuse.
The value on the x-axis can be calculated using the sine of the angle, like this:
sin 30 ° = x / 75
x = sin 30 ° * 75
x = 37.5
Now to calculate the value on the y-axis it would be by means of Pythagoras and it would be:
y ^ 2 = 75 ^ 2 - 37.5 ^ 2
y ^ 2 = 4218.75
y = 64.95, -64.95 because it must be located in the fourth quadrant
Which means that to draw the vector that we are asked to do, we must draw a point from origin (0,0) to (37.5, -64.95).
Then I leave a sketch of what it would be:
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Answer:
With the first four questions, all you need to do is find a value of x that gives you a zero value in any one of those brackets. Let's do it.
You said you finished 1, so let's skip it.
I can't give you a full answer for question 2, because you didn't capture the entire expression in the screenshot. I can see that the values 3 and 0.125 would give zero values, but I can't tell you the last.
For question three, we need to find x values where either 9x + 3 = 0, or x² - 9 = 0. There are three values that meet those conditions. You can solve it just by assigning a zero value and solving for x:
For the second term actually has two solutions, because a squaring a number also causes negatives to be converted to positives. Another way of seeing this is to notice that the second term is actually a difference of squares:
so in this case x could be -1/3, -3, or 3 to give a zero value.
For question four, you can only get a zero if you use imaginary numbers, as no real number can be squared to give you a negative. So the only answers to that are 5i, or -5i.
With questions five six seven and eight, we just need to make exceptions for zero denominators, so:
For question 5, you can x can be any real number except -4, 2 and 7
For 6, x can be anything but 3, 1/8th or 5
for 7, We again have that difference of squares, so the actual denominator is (9x + 3)(x + 9)(x - 9), meaning x can be anything but -1/3, -9, or 9
And finally for question 8, we again have a squaring, meaning that only an imaginary number will cause division by zero, meaning x can't be 5i or -5i.