Can someone please help me!!! I need help, no links !!!

Which expressions are equivalent?

timama [110]

Answer:

the anwser is b

Step-by-step explanation:

4 0

3 years ago

!!ONLY TWO QUESTIONS!!

Romashka-Z-Leto [24]

Answer:

A differnece between the plants is that the vascular plant has vascular vessels to carry water and food to all the different parts of the plant

Step-by-step explanation:

one of my teachers told me this.

3 0

3 years ago

Increase- from 42 acres to 72 acres

34kurt

42 acres to 72 acres, there is an increase of 30 acres
percentage wise
72-42=30
30/72=.4166...
therefore there is also a 41.6% increase

3 0

3 years ago

Read 2 more answers

Find the fifth roots of 243(cos 260° + i sin 260°).

Dmitry [639]

Answer:

z1=3 cos (52 + i sin 52)

z2 =3 cos (124 + i sin 124)

z3 = 3 cos (196 + i sin 196)

z4 =3 cos (268 + i sin 268)

z5= 3 cos (340 + i sin 340)

Step-by-step explanation:

To find the fifth roots of 243 (cos 260° + i sin 260°).

z ^ 1/5 = r^1/5 ( cis ( theta + 360 *k)/5) where k=0,1,2,3,4

So the first root of 243 (cos 260° + i sin 260°)

is z1 = 243^1/5 ( cis ( 260 + 360 *0)/5)

3 cis ( 260/5)

= 3 cis (52)

= 3 cos (52 + i sin 52)

The second root of 243 (cos 260° + i sin 260°)

is z2 = 243^1/5 ( cis ( 260 + 360 *1)/5)

3 cis ( 620/5)

= 3 cis (124)

= 3 cos (124 + i sin 124)

The third root of 243 (cos 260° + i sin 260°)

is z3 = 243^1/5 ( cis ( 260 + 360 *2)/5)

3 cis ( 980/5)

= 3 cis (196)

= 3 cos (196 + i sin 196)

The fourth root of 243 (cos 260° + i sin 260°)

is z4 = 243^1/5 ( cis ( 260 + 360 *3)/5)

3 cis ( 1340/5)

= 3 cis (268)

= 3 cos (268 + i sin 268)

The fifth root of 243 (cos 260° + i sin 260°)

is z5 = 243^1/5 ( cis ( 260 + 360 *4)/5)

3 cis ( 1700/5)

= 3 cis (340)

= 3 cos (340 + i sin 340)

6 0

3 years ago

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

Paha777 [63]

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,

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

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

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

We know that,

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

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

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

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

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

Using the trigonometric ratios, we get

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

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

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

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

6 0

3 years ago