What is another way to make 57 cents with coins

Mathematics

1 answer:

Fed [463]2 years ago

7 0

2 quarters, 7 pennies
5 dimes, 7 pennies
10 nickels, 7 pennies

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The number that is 10 less than 6

almond37 [142]

The number (x) that is (=) 10 less than (-) 6

x = 6 - 10 Subtract
x = -4

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2 years ago

How do I simplify this (sinx)/(secx+1)

pochemuha

Answer:

Step-by-step explanation:

$\frac{sin~x}{sec~x+1} =\frac{sin~x}{\frac{1}{cos~x}+1 } =\frac{sin~x~cos~x}{1+cos~x} \\=\frac{sin~x~cos~x}{1+cos~x} \times \frac{1-cos~x}{1-cos~x} \\=\frac{sin~x~cos~x(1-cos~x)}{1-cos^2 x} \\=\frac{sin~x~cos~x(1-cos~x)}{sin^2x} \\=\frac{cos~x(1-cos~x)}{sin~x} \\=cos ~x(csc~x-cot~x)\\or\\=cot~x(1-cos~x)$

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Which ordered pairs make the equation true?

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In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa

Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

$^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434$

Hence, the required number of ways =27434

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2 years ago

If you roll a six-sided die 1,000 times, about how many times would you expect to land on a three?

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