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PilotLPTM [1.2K]

3 years ago

10

Given: cos θ > 0, csc θ < 0

1 answer:

Greeley [361]3 years ago

4 0

Quadrant IV
Step-by-step explanation:
Use the mnemonics CAST.
The Cosine ratio and its reciprocal are positive in the fourth quadrant.
All the other trigonometric ratios are negative in the fourth quadrant.

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A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form

Over [174]

Answer:

$\vec{v}= \text{ or } \approx$

Step-by-step explanation:

Component form of a vector is given by $\vec{v}=$ , where $i$ represents change in x-value and $j$ represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector $\vec{v}=$ , the magnitude is $||v||=\sqrt{i^2+j^2$ .

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being $i$ , one leg being $j$ , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

$\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}$

To find the other leg, $i$ , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

$\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}$

Verify that $(9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark$

Therefore, the component form of this vector is $\vec{v}=\boxed{}\approx \boxed{}$

6 0

3 years ago

Help needed plaese✌

monitta

5x-3=2.5x+x solve this equation that I just gave you for X then plug it in to 2.5x and that will equal MN

7 0

3 years ago

Standard notation 8.15×10^5

sukhopar [10]

Answer:

8.15×10^5

Step-by-step explanation:

8.15×10^5

8.15×100000

815000

8 0

3 years ago

How would you write y=2/3x+1/3 in point slope form??

gladu [14]

Answer:

y - 1 = 2/3(x - 1).

Step-by-step explanation:

Point slope form is y - y1 = m(x - x1) where m = the slope and (x1, y1) is a point on the line.

y = 2/3x + 1/3 which gives the slope m = 2/3.

We can now find a point on the line by putting x = 1 say then find y:

y = 2/3(1) + 1/3 = 2/3 + 1/3 = 1 so the point is (1, 1).

(x1, y1) = (1, 1)

So the answer is ;

y - 1 = 2/3(x - 1).

6 0

3 years ago

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0

3 years ago