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Arada [10]
4 years ago
6

The values of trigonometric functions at multiples of 45 degrees are based on components of _____.

Mathematics
1 answer:
anygoal [31]4 years ago
6 0
<span>c. isosceles right triangles</span>
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What is the slope OF THE GRAPH, not what is A slope, what is THE slope
PtichkaEL [24]

You can find the slope either by just looking at the line or using the slope formula.

#1: The slope formula is:

m=\frac{y_2-y_1}{x_2-x_1}       Find two points and plug it into the formula

I will use (0, 2) and (1, -1)

(0, 2) = (x₁, y₁)

(1, -1) = (x₂, y₂)

m = \frac{y_2-y_1}{x_2-x_1}

m = \frac{2-(-1)}{0-1}   [two negatives cancel each other out and become positive]

m=\frac{2+1}{-1} =\frac{3}{-1}

m = -3

#2: To find the slope without having to do the work, you use this:

m=\frac{rise}{run}

Rise is the number of units you go up(+) or down(-) from each point

Run is the number of units you go to the right from each point

If we start at a defined/obvious point, like (0, 2), find the next point and see how many units it goes up or down and to the right. The next point is (1, -1), so from each point, you go down 3 units and to the right 1 unit. So your slope is -3/1 or -3. You can make sure the slope is right by looking at another point.

5 0
3 years ago
Read 2 more answers
Izara built a fence for her horses. The fence was 210 feet long. She put a fence post in the ground at the start of the fence an
Dmitry_Shevchenko [17]
B. $518.50 is the answer. Quizzes and Quizlet confirmed.
3 0
3 years ago
In a class of 25 students, 20 have a brother and 8 have a sister. There are 3 students
kaheart [24]

Answer: 6/8 or 3/4 or 0.75

Step-by-step explanation:

6 0
3 years ago
A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v&gt;=2i-4tj^
guapka [62]

Given :

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj  .

To Find :

A. The vector position of the particle at any time t .

B. The acceleration of the particle at any time t .

Solution :

A )

Position of vector v is given by :

d=\int\limits {v} \, dt\\\\d=\int\limits {(2i-4tj)} \, dt \\\\d=(2t)i+\dfrac{4t^2}{2}j\\\\d=(2t)i+(2t^2)j

B )

Acceleration a is given by :

a=\dfrac{dv}{dt}\\\\a=\dfrac{2i-4tj}{dt}\\\\a=\dfrac{2i}{dt}-\dfrac{4tj}{dt}\\\\a=0-4j\\\\a=-4j

Hence , this is the required solution .

5 0
3 years ago
21
Tju [1.3M]

Answer:

64x^127

Step-by-step explanation:

caculate it with all the numb

3 0
2 years ago
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