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notsponge [240]
3 years ago
12

Is 2,-5,-12,-19,-26 an arithmetic sequence

Mathematics
2 answers:
Mariulka [41]3 years ago
8 0
Yes!
-5 - 2 = -7;
-12 - (-5) = -7;
-19 -(-12) = -7;
-26 -(-19) = -7.
ExtremeBDS [4]3 years ago
4 0

Answer:

subtract 7 each time

Step-by-step explanation:

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Simplify the trigonometric expression.
Natasha_Volkova [10]
First we are going to find the common denominator of both fractions. To do that, we are going to multiply their denominators:
(1+sin \alpha )(1-sin \alpha )=1-sin^2 \alpha

Now we can rewrite our expression using the common denominator:
\frac{1-sin \alpha }{1-sin^2 \alpha } + \frac{1+sin \alpha }{1-sin^2 \alpha} = \frac{2}{1-sin^2 \alpha}

Finally, we can use the trig identities: 1-sin^2 \alpha =cos^2 \alpha and sec \alpha = \frac{1}{cos \alpha } to simplify our trig expression:
\frac{2}{cos^2\alpha}=2sec^2 \alpha

We can conclude that the correct answer is the fourth one.
5 0
3 years ago
Read 2 more answers
6x/5 - x = x/15 - 10/3
Vadim26 [7]

Answer:

x=-25

6x/5 - x = x/15 - 10/3

x/5 = x/15 - 10/3

3x = x - 50

3x - x = -50

2x = -50

x = -50/2

X=-25

4 0
3 years ago
Read 2 more answers
What is the LCD of the following rational expressions?
Aloiza [94]
\dfrac{x}{x+1} \ , \  \dfrac{7x}{x-1}

\dfrac{x (x -1)}{(x+1)(x -1)} \ , \  \dfrac{7x(x+1)}{(x-1)(x+ 1)}

<u>We know that (a + b)(a - b) = a² - b²:</u>

\dfrac{x (x -1)}{x^2 - 1}} \ , \ \dfrac{7x(x+1)} {x^2 - 1}

Answer: LCD = ( x + 1) (x - 1)
8 0
3 years ago
Read 2 more answers
Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select
Blizzard [7]

Answer:

Probability is:   $ \frac{\textbf{13}}{\textbf{51}} $

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $ \frac{number \hspace{1mm} of  \hspace{1mm} black  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm} of  \hspace{1mm} cards} $  $ = \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}} $

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $ \frac{number  \hspace{1mm} of  \hspace{1mm} red  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm}of  \hspace{1mm} cards} $  $ = \frac{26}{51}  $

Now, the probability of both black and red card = $ \frac{1}{2} \times \frac{26}{51} $

$ = \frac{\textbf{13}}{\textbf{51}} $

Hence, the answer.

5 0
3 years ago
Sorry for tabs!! I need help in this. Please help me.
telo118 [61]

Answer: 11 more i think

Step-by-step explanation:

8 0
3 years ago
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