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Naya [18.7K]
2 years ago
5

Help is it a b c or d

Mathematics
1 answer:
Dmitry [639]2 years ago
6 0

Answer:

<em>The last one) </em>

6 21/64 in.3

Step-by-step explanation:

Simply multiply LxWxH, (length x width x height.)

LxWxH= 3 1/2in x 1 7/8 in x 2 3/4 in= 6 21/64 in.3.

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balu736 [363]
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If a line contains the point (0, -1) and has a slope if 2, then which if the following points also lies on the line? (2, 1), (1,
Neko [114]
(1, 1) the slope is 2 and slopes can be read the easiest when expressed as a fraction (2/1) when the numerator (2) expresses how many digits you move the point up and the denominator (1) expresses how many points you move right
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2 years ago
Please help
katrin2010 [14]
\dfrac{n^2+3n+2}{n^2+6n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n^2+2n+n+2}{n^2+4n+2n+8}-\dfrac{2n}{n+4}\\\\=\dfrac{n(n+2)+1(n+2)}{n(n+4)+2(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{(n+2)(n+1)}{(n+2)(n+4)}-\dfrac{2n}{n+4}\\\\=\dfrac{n+1}{n+4}-\dfrac{2n}{n+4}=\dfrac{n+1-2n}{n+4}=\dfrac{1-n}{n+4}

Answer:\ \boxed{B.\ \dfrac{1-n}{n+4}}
7 0
3 years ago
PLEASE HELLPP!!!
Maslowich

Step-by-step explanation:

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4 0
3 years ago
there are n counters in a bag. 4 of the counters are red and the rest are blue. Ross takes a counter from the bag at random and
Ainat [17]

Answer: n = 10.

Step-by-step explanation:

In the bag, we have n counters.

4 of the counters are red.

the rest are blue, then we have:

(n - 4) blue counters.

Now, the probability that Ross takes a blue counter from the bag is equal to the quotient between the number of blue counters (n - 4) and the total number of counters, n

Then the probability is:

p1 = (n - 4)/n

Now he draws another, and it must be blue again, then we can calculate the probability in the same way as above, but he already take a blue counter, so the number of blue counters is (n - 5) and the total number of counters is (n - 1)

The probability of this event is:

p2 = (n - 5)/(n - 1)

The joint probability (the probability that Ross takes two blue counters) is equal to the product of the individual probabilities, and we know that this is equal to 1/3, then we have the equation:

1/3 = ( (n - 4)/n)*((n - 5)/(n - 1))

Now let's solve this for n.

n*(n - 1)/3 = (n - 4)*(n - 5)

(n^2 - n)/3 = n^2 - 4*n - 5*n + 20

n^2 - n = 3*(n^2 - 9*n + 20)

n^2 - n = 3*n^2 - 27*n + 60

0 = (3*n^2 - n^2) - 27*n + n + 60

0 = 2*n^2 - 26*n + 60

The two solutions of this equation can be found with Bhaskara's equation:

n = \frac{-(-26) +- \sqrt{(-26)^2 - 4*2*60} }{2*2} = \frac{26+- 14}{4}

Then the two solutions are:

n = (26 - 14)/4 = 3

This is not an option, because we know for sure that we have 4 red counters, then this option can be discarded.

The other solution is:

n = (26 + 14)/4 = 40/4 = 10

Then we have n = 10, 10 counters in total.

4 0
2 years ago
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