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kkurt [141]
3 years ago
13

Is the dolphin deeper than point C or point E?​

Mathematics
1 answer:
Pepsi [2]3 years ago
7 0
Point E is correct answer
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Prove this trigonometric identity<br>sin ​3x/sin x - cos 3x/cos x =2
Sveta_85 [38]

Step-by-step explanation:

LHS= sin 3x/sin x - cos 3x/cosx

Taking LCM,

<u>=</u><u>sin3xcosx- cos3xsinx</u>

sinxcosx

<u>=</u><u>sin(3x-x)</u>

sinxcosx

= <u>2sin2x</u>

2sinxcosx

=<u> </u><u>2sin2x</u><u> </u>

sin2x

=2

= RHS.

Proved.

8 0
3 years ago
What is 5c2×(3/4)2×(1/4)5-2
Jobisdone [24]
Very hard question....
7 0
3 years ago
Write in word form 64,307.009
Gnoma [55]

Answer:

sixty-four thousand three hundred seven and nine thousandths.

Step-by-step explanation:

3 0
3 years ago
What are the vertical asymptotes of the function f(x) = 2x + 8 / x2 + 5x + 6?
Irina18 [472]
I'm assuming the function is f(x) = (2x+8)/(x^2+5x+6). If so, make sure to use parenthesis to indicate that you're dividing all of "2x+8" over all of "x^2+5x+6" as one big fraction. Otherwise, things are ambiguous and it leads to confusion.

Side Note: x^2 means "x squared"

Factor the numerator: 2x+8 = 2(x+4)
Factor the denominator: x^2+5x+6 = (x+2)(x+3)

There are no common factors between the numerator and denominator. So there is nothing to cancel out. 

Recall that you cannot divide by zero. Something like 1/0 is undefined. 
We need to find the x values that cause the denominator to be zero.
Set the denominator equal to zero and solve for x
x^2+5x+6 = 0
(x+2)(x+3) = 0
x+2 = 0 or x+3 = 0
x = -2 or x = -3

The x values x = -2 or x = -3 will lead to the denominator being zero. This means that the vertical asymptotes are x = -2 or x = -3 as shown by the blue dashed vertical lines in the attached image.

4 0
3 years ago
When f(x) = x2+9 and g(x) = x-9, find f(x)-g(x).
ohaa [14]

Answer:

\huge{ \boxed{ \sf{ {x}^{2}  - x + 18}}}

Step-by-step explanation:

\underline{ \text{Given}}:

\sf{f(x) =  {x}^{2} +9  \: and \:  g(x) = x-9}

\underline{ \text{To \: find}} :  \sf{ \: f(x) \:  -  \: g(x)}

Put the values:

\mapsto{ \sf{ {x}^{2}  + 9 - ( x  - 9)}}

While subtracting , sign of each term of the second expression changes.

\mapsto{ \sf{ {x}^{2}  + 9 - x + 9}}

Add the numbers : 9 and 9

\mapsto{ \boxed{ \sf{{x}^{2}  - x + 18}}}

Hope I helped!

Best regards! :D

~\text{TheAnimeGirl}

5 0
3 years ago
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