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BARSIC [14]
3 years ago
7

(tan theta+sec theta-1)/(tan theta-sec theta+1)=sec theta+tan theta

Mathematics
1 answer:
9966 [12]3 years ago
8 0
What is the question
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Which of the numbers is rational? A) π B) 25 C) 2 D) 3
ivann1987 [24]
25, 2, and 3 are rational.
\pi is irrational.
3 0
3 years ago
Read 2 more answers
The following parts refer to the letters: LAMEFIREALARM. Recall that "word" means distinguishable letter arrangements. (a) How m
larisa86 [58]

Answer:

Step-by-step explanation:

The word "LAMEFIREALARM" word consists of

2L , 2M, 2E , 2R , 3A , F and I

no of words with the above letters such that M's are not together

=Total No of words- M's are together

Total no of words=\frac{13!}{4\times 2!\times 3!}=64,864,800

When M's are together . considering 2 M's as one

therefore there are \frac{12!}{3\times 2!\times 3!}=9,979,200

No of ways=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}=54,885,600

(b)No of ways such that M's are separated by at least 2 letters

at least  2 letter means 2 letter, 3 letter ......11 letters

So we have to subtract no of ways where there are 1 letter in between M's from total no of ways where 2 M's are not next to each other

No of ways in which there is 1 letter between 2 M's

This can be done by considering 11 cases

In first case Place first M in Starting Position and 2 M on third place

Second case place First M in 2 nd Position and second M on 4 th place

Similarly For 11th case

Place first M in 11th place and second M on 13th place

Total ways

=\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}

So , the total number of arrangements of the letters of the word LAMEFIREALARM where the two M's are separated by atleast two letters is

=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}-\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}

=45738000

7 0
3 years ago
Solve |p + 2| = 10Solve |p + 2| = 10<br><br> {-12}<br> {-8, 8}<br> {-12, 8}
valentinak56 [21]

By definition, we have

|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that p+2\geq 0 \iff p \geq -2

In this case, the absolute value doesn't do anything: the equation is

p+2 = 10 \iff p = 10-2 = 8

We are supposing p \geq -2, so we can accept this solution.

Now, suppose that p+2 < 0 \iff p < -2. Now the sign of the expression is flipped by the absolute value, and the equation becomes

-p-2 = 10 \iff -p = 12 \iff p = -12

Again, the solution is coherent with the assumption, so we can accept this value as well.

3 0
3 years ago
Which of the following is the decimal 0.6250 written as a fraction in lowest form
geniusboy [140]
The answer is 5/8. Hope this helps
6 0
3 years ago
SOLVE THE SIMULATENOUS EQUATIONS Y=X-2 AND Y= 3X+5
natta225 [31]

Answer: y = x-2

y = 3x + 5

x-2 = 3x+5

x+3x = 5+2

4x = 7

x = 7/4

y = x-2

y = 7/4 - 2

y = 7-8/4

y = -1/4

3 0
3 years ago
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