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vovikov84 [41]
3 years ago
14

(co|t^2x)/(1-sin^2x)

Mathematics
1 answer:
olga_2 [115]3 years ago
3 0
Hello :
cot(x) = cos(x)/sin(x)   .... cot²(x) = cos²(x)/sin²(x)
1-sin²(x) = cos²(x)
 cot²(x)(1-sin²(x)) = ( cos²(x)/sin²(x))× cos²(x) = 1/sin²(x)

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Help plz using rational exponents.
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Answer:

3. -3

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5. -5

Step-by-step explanation:

I can’t write radicals but fraction exponents change the number into a radical. Numerator being the power of the number and denominator being the type of root it is.

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Can someone help me on number 6?
MaRussiya [10]

Answer:

b=72

Step-by-step explanation:

1/3(b+6) = 1/4b+8

First, open the parentheses, which means multiplying both variables in the parentheses by 1/3.

1/3b+2 = 1/4b+8

Then, subtract 2 from both sides.

1/3b = 1/4b+6

Subtract 1/4b from both sides, but first change the fractions so that they have the same denominator.

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3 years ago
About 2/3 of Mrs. Lyons students have two or more pets. If she has 98 students, about how many of her students have LESS than tw
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Step-by-step explanation:

Since 2/3 of her students have two or more pets, this means....

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You roll a die with the sample space s=1,2,3,4,5,6. you define A as (1,2,4) B as (2,3,4,5,6) C as (3,4) and D as (2,4,5). Determ
bekas [8.4K]

Answer:

A and B are exhaustive.

Step-by-step explanation:

Given

A = \{1,2,4\}

B = \{2,3,4,5,6\}

C =\{3,4\}

D = \{2,4,5\}

Solving (a): The mutually exclusive events

These are events that have no common or mutual elements

Events A to D are not mutually exclusive because each of the events have at least 1 common element with one another.

Solving (b): Exhaustive events.

Two events X and Y are said to be exhaustive if:

S = P(X\ n\ Y)

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For events A to D, we have:

A\ n\ B = \{1,2,3,4,5,6\}

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By comparison;

A\ n\ B = S

Hence, A and B are exhaustive.

8 0
3 years ago
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