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2 years ago

Which of the following is an equation of a line that is parallel to one whose equation is y = 5x - 3?

Mathematics

1 answer:

LekaFEV [45]2 years ago

3 0

Any line that has 5 as its slope. So which ever one says “5x”

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Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago

PLs help me on this..........

Serggg [28]

Answer:

247277 people

Step-by-step explanation:

6% of 243,000 = 14,580

243,000 + 14,580 = 257,580

4% of 257,580 = 10,303.2

Estimating 10,303.2 to 10,303

257,580 - 10,303 = 247277

3 0

2 years ago

Read 2 more answers

Write a verbal phrase for the following algebraic expression 12y

aleksandrvk [35]

The product of a number y and 12 or 12 times a number y. These are both equivalent expressions to 12y

6 0

2 years ago

Tanx cosx = 0

shtirl [24]

You're almost finished.

(sin/cos) times cos = 0

Look at the left side. You could write it as (sin x cos) / cos = 0
and simply divide numerator and denominator by the cosine (cancel it).

Then what do you have left ? . . . sin(x) = 0 Do I need to finish this for you ?

5 0

3 years ago

Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k

valkas [14]

Answer:

$\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k, the vector projection of B unto a is expressed as $proj_ab = \dfrac{b.a}{||a||^2} * a$

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||² = 130

$proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\$

Hence the projection of b unto a is expressed as $\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\$

7 0

3 years ago