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natta225 [31]
3 years ago
15

What is the value of h?

Mathematics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

I don't know what to approximate to, but the exact answer is 100√3, with 173.2050808 being the decimal answer.

Step-by-step explanation:

Use the sin function:

sin\alpha = \frac{opposite}{hypotenuse}

sin60 = \frac{h}{200}

h = 200sin60

h = 200(\frac{\sqrt{3}}{2} ) = 100\sqrt{3} = 173.2050808

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Can someone explain/show me how to do this?Absolute Value Functions
dedylja [7]
Domain: -∞<x<∞
Range: -∞<x<∞
X-Intercept: x=0
Y-Intercept: y=0
Increasing on the interval of 0<x<∞
<span>Decreasing on the interval of -∞<x<0


</span>When A=0, the graph equals y=0
- When A is greater than 1, it makes the graph skinnier than <span>f(x)=|x|
- When A is less than 1 but greater than 0, it makes the graph fatter than </span><span>f(x)=|x|
- When A turns negative, it flips the graph upside down. 

-When B is greater than 0, it translates the graph to the right
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When C is greater than 0, the graph moves upwards
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3 0
3 years ago
Jacob scored 85%, 90%, and 74% on his first three tests. Then he scored 99%, 84% and 75% on his next three
AlekseyPX
The average of the first three is 83 (85 + 90 + 74)/3
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3 0
3 years ago
Which set of ordered pairs does not represent a function? \{(5, -9), (6, -6), (-3, 8), (9, -6)\}{(5,−9),(6,−6),(−3,8),(9,−6)} \{
Nady [450]

Answer:

\{(-6, -4), (4, -8), (-6, 9), (1, -3)\}

Step-by-step explanation:

Given

\{(5, -9), (6, -6), (-3, 8), (9, -6)\}

\{(-6, -4), (4, -8), (-6, 9), (1, -3)\}

\{(1, -1), (-5, 7), (4, -9), (-9, 7)\}

\{(8, -9), (-3, -6), (-4, 4), (1, -5)\}

Required

Which is not a function

An ordered pair is represented as:

\{(x_1,y_1),(x_2,y_2),(x_3,y_3),..........,(x_n,y_n)\}

However, for the ordered pair to be a function; all the x values must be unique (i.e. not repeated)

<em>From options (a) to (d), option (b) has -6 repeated twice. Hence, it is not a function.</em>

4 0
3 years ago
If a number is not a natural number then it is not a whole number
Alisiya [41]

Answer:

no

Step-by-step explanation:

a natural number has to be a whole number

4 0
3 years ago
Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
3 0
4 years ago
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