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oksano4ka [1.4K]
2 years ago
7

The number of candies in the bag was less than 80. If it is possible to divide all the candies evenly among 3, 4, and 5 children

, how many candies were in the box?
Mathematics
2 answers:
Lelechka [254]2 years ago
6 0

Answer:

60 candies

Step-by-step explanation:

multiply 3 by 4 by 5  and you get 60, which has 3, 4, and 5 as prime factor, thus, making it possible to have an equal share with less than 80 candies

insens350 [35]2 years ago
4 0

60 candies

3*4*5=60

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is the degree of ( f+g)(x) equal to the degree of f(x) or g(x) , which ever has the highest degree ?​
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Answer:

for all x in the domain of f(x), or odd if, f(−x) = −x, for all x in the domain of f(x), or neither even nor odd if neither of the above are true statements. A kth degree polynomial, p(x), is said to have even degree if k is an even number and odd degree if k is an odd number

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3 years ago
Please help me guys it is due today there is a second part i will do anything so ANSWER PLS!!! ASAP
Fantom [35]

Answer:

see below

Step-by-step explanation:

To find the combined length of Pine and Elm and the lengths together

Pine +Elm

We are estimating by rounding to the nearest whole number

Pine is 2.7 so it rounds to 3 because we look at the .7 and since .7 is greater than .5 it rounds the 2 to 3

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5 0
2 years ago
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Molodets [167]
16 nickles
10 dimes

1.00$ + 0.80$
5 0
2 years ago
Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!
LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}

=\sqrt{40}b^2\sqrt{24b^3}

\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

=\sqrt{24}b^{\frac{3}{2}}

=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

4 0
3 years ago
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