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3 years ago

Can some one please help me with my math and tell me if it’s correct thanks

Mathematics

1 answer:

crimeas [40]3 years ago

8 0

Yes, looks very correct.

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Solve for b. Round your answer to the nearest whole degree.

Talja [164]

Answer:

44°

Step-by-step explanation:

From sin rule:

a / sin A = b / sin B = c / Sin C

3.6 / sin b = 5.1 / sin 100

Cross multiply :

5.1 * sinb = 3.6 * sin 100

Sin b = 3.5453079 / 5.1

Sin b = 0.6951584

b = sin^-1(0.6951584)

b = 44.039

b = 44°

8 0

3 years ago

What is the ratio of the width to the length of a rectangular field that has a perimeter of 120 meters and an area of 500 square

Vinil7 [7]

Answer:

Step-by-step explanation:

w=50m

10m

6 0

3 years ago

941 rounded to the nearest 10

Oksana_A [137]

Answer:

900

Step-by-step explanation:

you just see if by the number, if the number is 6 and up your round over and it would become 950 but the number is lower so you make it 900.

3 0

3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

How do I solve this???

olga55 [171]

Answer:

by using the graph!

Step-by-step explanation:

8 0

3 years ago