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3 years ago

Can someone help with this and explain answer please.

Mathematics

2 answers:

Nikolay [14]3 years ago

6 0

The answer should be 18 as they are similar triangles.
So I did this.
I took the 24/52 as the are similar triangles
The took that decimal and multiplyed it by 39 to get 18

photoshop1234 [79]3 years ago

3 0

Because the triangles are similar the ratio of all the sides would be the same. So you need to find the ratio of the common sides and then use that ratio to find the unkown side.

In this picture sides FG = 52, FV would be the same side and is 24.

Dividing 24/52 = 0.461, so the unkown side WV would be 0.444 of it's common side HG.

HG = 39

39 * 0.461= 18.

The answer is C. 18

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which of the following best describes 4^2/3? a. sq root of 16, b. cube root of 4, c. sq root of four, d. cube root of 16

son4ous [18]

Answer: D) cube root of 16

================================================

Explanation:

The rule we use is

$x^{m/n} = \sqrt[n]{x^m}$

In this case, x = 4, m = 2 and n = 3.

So,

$x^{m/n} = \sqrt[n]{x^m}\\\\\\4^{2/3} = \sqrt[3]{4^2}\\\\\\4^{2/3} = \sqrt[3]{16}\\\\\\$

Showing that the original expression turns into the cube root of 16.

4 0

3 years ago

What is the difference between categorical and quantitative data? Give examples of each of your own for each. Also decide whethe

Snezhnost [94]

Categorical data may or may not have some logical order while the values of a quantitative variable can be ordered and measured.

Categorical data examples are: race, sex, age group, and educational level

Quantitative data examples are: heights of players on a football team; number of cars in each row of a parking lot

a) Colors of phone cover - quantitative

b) Weight of different phones - quantitative

c) Types of dogs - categorical

d) Temperatures in the U.S. cities - quantitative

6 0

3 years ago

Someone help me please!!!!

Ghella [55]

I am pretty sure number 3 is the right answer

8 0

3 years ago

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QveST [7]

If I’m correct it should be 7 because range is least greatest subtracted from most greatest and the problem is asking from the green box so 67-60=7

8 0

3 years ago

Read 2 more answers

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