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serious [3.7K]
1 year ago
8

Use the drop-down menus to complete the solution to the equation cosine (startfraction pi over 2 endfraction minus x) = startfra

ction startroot 3 endroot over 2 endfraction for all possible values of x on the interval [0, 2pi].
Mathematics
1 answer:
Lilit [14]1 year ago
6 0

Use the drop-down menus to complete the solution to the equation cosine (start fraction pi over 2 end fraction minus x) = start fraction start root 3 end root over 2 end fraction for all possible values of x on the interval [0, 2pi].

Using trigonometric identities, the solution to the equation cos(\frac{\pi }{2}-x) = \frac{\sqrt{3} }{2} for all possible values of x on the interval [0, 2π].

What are trigonometric identities?

Trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined.

cos(\frac{\pi }{2}-x) = \frac{\sqrt{3} }{2}  \\\\cos(x)cos\frac{\pi }{2}+sin(x)sin \frac{\pi }{2} = \frac{\sqrt{3} }{2}\\\\cos(x)(0)+sin(x)(1) = \frac{\sqrt{3} }{2}\\\\sin(x) = \frac{\sqrt{3} }{2}\\\\x = \frac{\pi }{3}, \frac{2\pi }{3}

To learn more about trigonometric identities click here brainly.com/question/7331447

#SPJ4

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Deshawn has two bags of marbles. The first bag has 2 blue, 3 orange, and 5 red. The second bag has 4 pink, 10 blue, and 6 brown.
madreJ [45]

Answer: 0.1

Step-by-step explanation:

Given

Bag-I has 2 blue,3 orange, 5 red

Bag-II has 4 Pink,10 blue, 6 brown

No of ways of choosing a blue marble from bag-I

\Rightarrow ^2C_1

Total no of ways of choosing a marble from bag-I

\Rightarrow ^{10}C_1

No of ways of choosing a blue marble from bag-II

\Rightarrow ^{10}C_1

Total no of ways of choosing a marble from bag-II

\Rightarrow ^{20}C_1

The probability that he will pull out a blue marble from each bag is

\Rightarrow P=\text{Probability of pulling a blue marble from bag-I}\times \text{Probability of pulling a blue out bag-II}

\Rightarrow P=\dfrac{^2C_1}{^{10}C_1}\times \dfrac{^{10}C_1}{^{20}C_1}\\\\\Rightarrow P=\dfrac{2}{10}\times \dfrac{10}{20}=\dfrac{1}{10}

3 0
3 years ago
Is 2.12359..... rational???​
den301095 [7]

Answer:

No

Step-by-step explanation:

Rational numbers are those that can be represented in a fraction form.

2.12359.... cannot be represented in a fraction form.

So, it is NOT a rational number

6 0
3 years ago
Read 2 more answers
NEED HELP ASAP PLEASE
timurjin [86]

If the unknown value is x, then x = \frac{14}{4} × 6 = 21.

Explanation:

  • Since it is a double number line with a ratio, all the ratios of one number line to the other will remain equal for all the values along both the number lines. Assume the unknown value to be x.
  • The ratio of the number on the first number line to the corresponding number on the second number line is calculated to be;                                  \frac{7}{2} = \frac{14}{4} = \frac{x}{6} = \frac{28}{8} = \frac{35}{10} = 3.5. So \frac{x}{6} = 3.5, x = 3.5 × 6 = 21.
  • We can also calculate x by the following method. As ratios are equal we can directly substitute the value to another ration.                                                      So \frac{14}{4} = \frac{x}{6}, x = \frac{14}{4} × 6 = \frac{84}{4} = 21.

3 0
2 years ago
Read 2 more answers
Match the numerical expressions to their simplest forms.
Aloiza [94]

Answer:

(a^6b^1^2)^\frac{1}{3} = a^2b^4

\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}} = a^3b^2

(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4} = a^2b

(\frac{a^3}{ab^-^6})^\frac{1}{2} = ab^3

Step-by-step explanation:

Simplify each of the expressions:

1

(a^6b^1^2)^\frac{1}{3}

Distribute the exponent. Multiply the exponent of the term outside of the parenthesis by the exponents of the variable.

(a^6b^1^2)^\frac{1}{3}

a^6^*^\frac{1}{3}b^1^2^*^\frac{1}{3}

Simplify,

a^2b^4

2

Use a similar technique to solve this problem. Remember, a fractional exponent is the same as a radical, if the denominator is (2), then the operation is taking the square root of the number.

\frac{(a^5b^3)^\frac{1}{2}}{(ab)^-^\frac{1}{2}}

Rewrite as square roots:

\frac{\sqrt{a^5b^3}}{\sqrt{(ab)}^-^1}

A negative exponent indicates one needs to take the reciprocal of the number. Apply this here:

\frac{\sqrt{a^5b^3}}{\frac{1}{\sqrt{ab}}}

Simplify,

\sqrt{a^5b^3}*\sqrt{ab}

Since both numbers are under a radical, one can rewrite them such that they are under the same radical,

\sqrt{a^5b^3*ab}

Simplify,

\sqrt{a^6b^4}

Since this operation is taking the square root, divide the exponents in half to do this operation:

a^3b^2

3

(\frac{a^5}{a^-^3b^-^4})^\frac{1}{4}

Simplify, to simplify the expression in the numerator and the denominator, the base must be the same. Remember, the base is the number that is being raised to the exponent. One subtracts the exponent of the number in the denominator from the exponent of the like base in the numerator. This only works if all terms in both the numerator and the denominator have the operation of multiplication between them:

(\frac{a^8}{b^-^4})^\frac{1}{4}

Bring the negative exponent to the numerator. Change the sign of the exponent and rewrite it in the numerator,

(a^8b^4)^\frac{1}{4}

This expression to the power of the one forth. This is the same as taking the quartic root of the expression. Rewrite the expression with such,

\sqrt[4]{a^8b^4}

SImplify, divide the exponents by (4) to simulate taking the quartic root,

a^2b

4

(\frac{a^3}{ab^-^6})^\frac{1}{2}

Using all of the rules mentioned above, simplify the fraction. The only operation happening between the numbers in both the numerator and the denominator is multiplication. Therefore, one can subtract the exponents of the terms with the like base. The term in the denomaintor can be rewritten in the numerator with its exponent times negative (1).

(a^3^-^1b^(^-^6^*^(^-^1^)^))^\frac{1}{2}

(a^2b^6)^\frac{1}{2}

Rewrite to the half-power as a square root,

\sqrt{a^2b^6}

Simplify, divide all of the exponents by (2),

ab^3

7 0
3 years ago
$384 with a discount of 6.5%​
andrey2020 [161]

Answer:

$359.04

Step-by-step explanation:

384 x .935 = 359.04

3 0
2 years ago
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