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BabaBlast [244]
3 years ago
14

What is the value of sin(A)?

Mathematics
2 answers:
Rama09 [41]3 years ago
6 0

Answer:

3/5

Step-by-step explanation:

nika2105 [10]3 years ago
5 0

Answer:

Sin(A) =3/5 is answer

Step-by-step explanation:

I hope it's helpful!

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Which formula can be used to describe the sequence?
alexdok [17]

Answer:

f(x+1) = -3/4 × f(x)

Step-by-step explanation:

first of all, the sign of the numbers in the sequence is alternating. so, there must be a "-" involved.

that eliminates the first and third answer options.

and the absolute values of the numbers in the sequence are going down. |f(x+1)| < |f(x)|

that eliminates the fourth answer option, as this says that

|f(x)| < |f(x+1)|. and that is the opposite of how the actual sequence behaves.

7 0
3 years ago
PLEASE HELP 100 POINTS<br> write an equation of the horizontal line that passes through (-7,-10)
Vedmedyk [2.9K]

Answer:

y + 10 = 0

Step-by-step explanation:

Slope of horizontal line is zero (0)

m = 0

By point slope form of a line.

y-y_1 =m(x - x_1) \\  \\ y - ( - 10) = 0 \{x - ( - 7) \} \\  \\ y + 10 = 0 \{x + 7\} \\  \\y + 10 = 0 \\  \\

5 0
2 years ago
Read 2 more answers
Can yall help me do this
4vir4ik [10]

Formula = 4/3 x pi x r^3

4/3 x 3.14 x 7^3 = 1436.03


answer: a. 1436.03 cubic units

3 0
2 years ago
Read 2 more answers
The radius of Earth is about 3,980 miles. What expression can be used to find the approximate diameter of Earth?
Tatiana [17]
To find the diameter, you use the formula: 2x. we use 2x because the radius is 2 times the diameter.

so, if the radius is 3,980, then we substitute that into the formula.

2(3,980)

this will give us 7960. the diameter is 7960 miles.
3 0
3 years ago
How do you find the limit?
coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}

We can cancel x-5.

\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}

Then directly substitute x = 5 in.

\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\&#10;&#10;\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}

<u>Differentiation</u> (Property of Addition/Subtraction)

\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}

Hence from the expressions,

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}

<u>Differential</u> (Constant)

\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \  is \ \ a \ \ constant.)}}

Therefore,

\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}

Now we can substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}

Thus, the limit value is 2/5 same as the first method.

Notes:

  • If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.
8 0
3 years ago
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