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crimeas [40]
3 years ago
7

Sonia saves

Mathematics
2 answers:
leonid [27]3 years ago
7 0
Sonia saved $594 for retirement
Y_Kistochka [10]3 years ago
5 0

Answer:

Sonia saved 1889 from her last month's pay

Step-by-step explanation:

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PLSSS HELP DUE IN 5 MINUTES!!!!
Pavel [41]

Answer:

I think it's (0,3) sorry if it not

4 0
2 years ago
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guys I really need help with part c) I literally have no idea how to this. given that tan(pi/8)=sqrt(2)-1. I am really looking f
damaskus [11]

Answer:

a \geq \frac{1}{\sqrt{2} -1}

Step-by-step explanation:

This equation is more intimidating than the problem you have to solve.


You know that the sine of everything is always between -1 and +1. So for the entire expression to be >= 0, the a*tan(pi/8) bit has to be 1 at least. Given this, we can forget about the sin(...) term of the equation for the remainder of solving it.



You already figured out that tan(pi/8) is sqrt(2)-1.



So what we're saying is a * (sqrt(2) - 1) has to be 1 at least.



If we solve a(sqrt(2)-1) >= 1 for a we get:



a = 1/(sqrt(2)-1)

6 0
3 years ago
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Amelia and Elliott are collecting empty soda cans for recycling. Amelia has 13 less than 5 times the cans that Elliott has. Toge
xeze [42]

Answer:

D.) Elliott has 212 cans, Amelia has 45 cans

Step-by-step explanation:

257 = x + (5x - 12)

257 = 6x - 13

+13           +13

270 = 6x

/6        /6

45 = x     (the amount of Amelia's cans)


257 - 45 = 212  (the amount of Elliott's cans)


please rate and thank : )

6 0
3 years ago
△PQR is reflected to form​ ​ △P′Q′R′ ​. The vertices of ​ △PQR ​ are P(1,1) , Q(−1,−2) , and R(4,−2) . The vertices of ​ △P′Q′R′
kozerog [31]
The answer is y=-x.
8 0
3 years ago
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2 cards are drawn from a standard deck without replacement. What is the probability that at least one of the cards drawn is an a
olya-2409 [2.1K]

Answer:

<h2>The answer is 0.1493.</h2>

Step-by-step explanation:

In a standard deck there are 52 cards in total and there are 4 aces.

Two cards can be drawn from the 52 cards in ^{52}C_2 = \frac{52!}{50!\times2!} = 51\times26 ways.

There are (52 - 4) = 48 cards rather than the aces.

From these 48 cards 2 cards can be drawn in ^{48}C_2 = \frac{48!}{46!\times2!} = 47\times24 ways.

The probability of choosing 2 cards without aces is \frac{47\times24}{51\times26} = \frac{188}{221}.

The probability of getting at least one of the cards will be an ace is 1 - \frac{188}{221} = \frac{33}{221} = 0.1493.

6 0
3 years ago
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