All categories

3 years ago

Can ya help me out ?? Thanks

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

5 0

Answer:

n=1

can you also please mark me as brainliest?

thanks!

You might be interested in

What is the rule of a function of the form f(t)= a sin (bt+c) +d whose graph appears to be identical to the given graph?

Gnoma [55]

Answer:

Option D.

Step-by-step explanation:

We can easily solve this problem by using a graphing calculator or plotting tool.

The function is

f(t) = a*sin (b*t +c) + d

Please, see attached picture below.

By looking at the picture with all the possible cases, we can tell that the correct option is D.

The function has a period of T = 6π

Max . Amplitude = 9

Min . Amplitude = -5

7 0

3 years ago

Read 2 more answers

Match each value with its formula for ABC.

MariettaO [177]

The solution to the question is:

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

<h3>What is cosine rule?</h3>

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

$c^{2}$ = $a^{2}$ + $b^{2}$ -2ab cos C-----------------1

then c = $\sqrt{a^{2} + b^{2} -2abcosC }$

if b is the side facing the included angle B, then

$b^{2}$ = $a^{2}$ + $c^{2}$ -2accosB-----------------2

b = $\sqrt{a^{2} + c^{2} -2accosB }$

from equation 2, make cosB the subject of equation

2ac cosB = $a^{2}$ + $c^{2}$ - $b^{2}$

cosB = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

if a is the side facing the included angle A, then

$a^{2}$ = $b^{2}$ + $c^{2}$ -2bccosA--------------------3

a = $\sqrt{b^{2} + c^{2} -2bccosA }$

from equation 3, making cosA subject of the equation

2bcosA = $b^{2}$ + $c^{2}$ - $a^{2}$

cosA = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

from equation 1, making cos C the subject

2abcosC = $b^{2}$ + $a^{2}$ - $c^{2}$

cos C = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

In conclusion,

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

Learn more about cosine rule: brainly.com/question/4372174

$SPJ1

4 0

2 years ago

What the square root 11?

Mumz [18]

I hope this helps you

5 0

3 years ago

How do I solve y=x+3

Lena [83]

You can use a <u>graph calculator</u>, if you don’t have one use the app <u>Photomath</u> to help.

4 0

3 years ago

Read 2 more answers

Determine which two values the following number is between. A. 12 and 13 B. 13 and 14 C. 15 and 16 D. 14 and 15

nordsb [41]

Answer:

more info pls

Step-by-step explanation:

8 0

2 years ago