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Evgen [1.6K]
3 years ago
11

a shop sells sweets in bags of 7 and 20 what is the largest number of sweets that cannot be purchased exactly?

Mathematics
1 answer:
bogdanovich [222]3 years ago
8 0
By the Frobenius Coin Theorem, also known as the Postage Stamp Theorem or the Chicken McNugget Theorem, the maximum amount that cannot be represented as a sum of multiples of two relatively prime numbers a and b (numbers that have GCF of 1) is ab-a-b.  Plugging 7 and 20 in, our answer is 7*20-7-20, or 140-27, or 113.
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HELP HELP HELP
Law Incorporation [45]

Based on her results, if she flipped the coins another 50 times, she should expect to flip heads 20 times. If Jessica flips a coin 100 times and gets 40 times heads and the next time she flips a coin 50 times she should get 20 because, 100 divide 2 is 50 so you would have to divide 40 with 2 and get the answer of 20.

3 0
3 years ago
Write the equation for function g(x)
iragen [17]

Answer:

g(x) = (x + 2)^2 + 1

Step-by-step explanation:

From the graph/image that you have provided said translated shift of

f(x) -> g(x)

f(x) = x^2 , g(x) = a(x -h)^2 +k

h is shift right/left

k is shift up/down.

It appears that the shift is left 2 and up 1.

h = -2 and k = 1

g(x) = (x - (-2))^2 +1

g(x) = (x + 2)^2 + 1

5 0
2 years ago
How many different four-digit numbers can be formed with the numbers 7; 4; 5; 1; 2; 9; and 8?
Masja [62]

Answer:

If I'm correct, the answer is 1296.

Step-by-step explanation:

6 x 6 x 6 x 6 = 1296

7 0
3 years ago
Find the area of each. Use your calculator's value of p. Round your answer to the nearest tenth.
Neko [114]

Answer:

22/7*11*11=38.285

this must be ans

8 0
3 years ago
Factor 7(x − 3)2 − 4(x − 3) − 3 completely.
Nezavi [6.7K]
7(x-3)^2-4(x-3)-3\\\\substitute\ t=x-3,\ then\ we\ have\\\\7t^2-4t-3\\\\a=7;\ b=-4;\ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot7\cdot(-3)=16+84=100\\\\t_1=\dfrac{-b-\sqrt\Delta}{2a};\ t_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{100}=10\\\\t_1=\dfrac{-(-4)-10}{2\cdot7}=\dfrac{4-10}{14}=\dfrac{-6}{14}=-\dfrac{3}{7}\\\\t_2=\dfrac{-(-4)+10}{2\cdot7}=\dfrac{4+10}{14}=\dfrac{14}{14}=1\\\\7t^2-4t-3=7\left(t+\dfrac{3}{7}\right)(t-1)=(7t+3)(t-1)\\\\=(7(x-3)+3)(x-3-1)=(7x-21+3)(x-4)\\\\=\boxed{(7x-18)(x-4)}
7 0
4 years ago
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