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2 years ago

What is the slope of this line A. -3/4 B. 4 C. -3 D. 3/4

Mathematics

1 answer:

djverab [1.8K]2 years ago

7 0

Your answer is a rise over run and the line is in a negative slope

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Sin(60-theta)sin(60+theta)

ehidna [41]

Step-by-step explanation:

$\sin(60 - \theta) \sin(60 + \theta) \\ = \{ \sin(60) \cos( \theta) - \sin( \theta) \cos(60) \} \{ \sin(60) \cos( \theta) + \cos(60) \sin( \theta) \} \\ = \{ \frac{ \sqrt{3} }{2} \cos( \theta) - \frac{1}{2} \sin( \theta) \} \{ \frac{ \sqrt{3} }{2} \cos( \theta) + \frac{1}{2} \sin( \theta) \} \\ \\$

from difference of two squares:

${ \boxed{(a - b)(a + b) = ( {a}^{2} - {b}^{2} ) }}$

therefore:

$= \{ {( \frac{ \sqrt{3} }{2}) }^{2} { \cos }^{2} \theta \} - \{ {( \frac{ \sqrt{3} }{2} )}^{2} { \sin }^{2} \theta \} \\ \\ = \frac{3}{4} { \cos }^{2} \theta - \frac{3}{4} { \sin}^{2} \theta$

factorise out ¾ :

$= \frac{3}{4} ( { \cos }^{2} \theta - { \sin}^{2} \theta) \\ \\ = { \boxed{ \frac{3}{4} \cos(2 \theta) }}$

3 0

3 years ago

Read 2 more answers

The value of a car depreciates by 35% per year. Work out the current value of a car bought 2 years ago for £20000

Rzqust [24]

First year: the depreciation is (35/100) x 20000 = £7000; now the value of the car is £20000 - £7000 = £13000;

Second year: the depreciation is (35/100) x 13000 = £4550; the current value of the car is £13000 - £4550 = £8450.

6 0

3 years ago

4 divided by 57 please help

Basile [38]

Answer:

0.07017

Step-by-step explanation:

6 0

2 years ago

A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

VMariaS [17]

Answer:

Q = arctan(7.1739) = 82.06

Step-by-step explanation:

Given:

- The mass of the person m = 20.3 kg

- The distance traveled up the ladder s = 1.1 m

- The gravitational constant g = 9.8 m/s^2

- The coefficient of static friction u_s = 0.23

- Total length of the ladder

Find:

The minimum angle θ, that would allow the person to climb without ladder slipping

Solution:

- Taking moments about point of ladder and wall contact A to be zero:

-F_n,b*cos(Q)*4 - F_f*sin(Q)*4+ m*g*cos(Q)*2.6 = 0

- Taking Sum of vertical forces to be zero:

F_n,b - m*g = 0

F_n,b = m*g

- The frictional force F_f is given by:

F_f = u_s*F_n,b = u_s*m*g

- Plug the values back in:

- m*g*cos(Q)*4 + u_s*m*g*sin(Q)*4 - m*g*cos(Q)*2.6 = 0

Simplify:

4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)

6.6*cos(Q) = 4*u_s*sin(Q)

tan(Q) = 6.6 / 4*u_s

- Plug in the values:

tan(Q) = 6.6 / 4*0.23

Q = arctan(7.1739) = 82.06

4 0

3 years ago

Evaluate 2/5g+3h-6 when g=10 and h=6

Reptile [31]

2/5(10)+3(6)-6
4+18-6
22-6
16
i hope this helps!

8 0

3 years ago

Read 2 more answers