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3 years ago

Find the value of x

Mathematics

1 answer:

balu736 [363]3 years ago

3 0

Answer:

x = 4

Step-by-step explanation:

This is an isosceles triangle since the base angles are congruent.

If we know it's an isosceles triangle then we know that the legs of it are congruent.

9x - 8 = 28

9x = 36

x = 4

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Trina, James, and Michael are training for a 10-kilometer race. After 30 minutes of a training run, Michael lead the group, and

Llana [10]

Answer:

1/3

Step-by-step explanation:

1/2 James' fraction of Michael

2/3 Trina's fraction of James

1/2 x 2/3 = 1/3

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3 years ago

The average weight of a medium size egg is found to be 3.2 oz, with a standard deviation of 0.13 oz. What is the heaviest egg (t

Over [174]

Answer:

We are 95% confident that the average weight of the heaviest egg that is still medium size is 3.52 oz

Step-by-step explanation:

Mean = 3.2 oz, sd = 0.13 oz

Assuming there are 3 heavy eggs, n= 3, degree of freedom = n-1 = 3-1 = 2, t-value corresponding to 2 degrees of freedom and 95% confidence level is 4.303

Error margin = t×sd/√n = 4.303×0.13/√3 = 0.32 oz

Upper limit (heaviest egg that is still medium size) = mean + error margin = 3.2 + 0.32 = 3.52 oz

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

A 500.-kilogram roller coaster starts from rest at the top of an 80.0-meter hill. What is its potential enegy halfway down the h

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Answer:

the answer is 200kg

Step-by-step explanation:

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3 years ago

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