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ycow [4]
3 years ago
15

Porter's bird feeder holds 1/2 of a cup of birdseed. Porter is filling the bird feeder with a scoop that holds 1/10 of a cup. Ho

w many scoops of birdseed will Porter put into the feeder?
Write your answer as a fraction or as a whole or mixed number.
Mathematics
1 answer:
Arlecino [84]3 years ago
3 0

Answer:

5 scoops

Step-by-step explanation:

If the feeder already has 1/2 cups in it, Porter will need to put 5 scoops in because 5/10= 1/2. 5 is half of ten

Hope this helps!

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Can someone teach me how to do this?
vladimir1956 [14]

the formula is a_{1}+(n-1) d

so a_{1} would be the first number in the sequence, which would be 13 in problem 9.

13+(n-1)d

then you put in n, which is 10 (it represents which number in the sequence you're looking for, for example 16 is the second number in the sequence)

13+(10-1)d

then you find the difference between each number, represented by d which in this case is 3

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3 years ago
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Evgen [1.6K]

Answer:

6

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qwelly [4]

Answer:

1) \dfrac{12+\sqrt{-16}}{4}=\dfrac{12+\sqrt{16i^{2}}}{4}\\\\\\=\dfrac{12+4i}{4}=\dfrac{12}{4}+\dfrac{4i}{4}\\\\\\= 3 + i

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2) 4 + 7i - 8 = 4 -8 + 7i

                   = -4 + 7i

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3)\dfrac{\sqrt{9}+\sqrt{-4}}{5}=\dfrac{3+\sqrt{4i^{2}}}{5}\\\\\\=\dfrac{3}{5}+\dfrac{2i}{5}\\\\\\Real \ part = \dfrac{3}{5} \ & \ imaginary \ part = \dfrac{2}{5}

4) -2-\sqrt{-25}=-2-\sqrt{25i^{2}}=-2-5i

Real part = -2  & imaginary part = -5

5)\dfrac{15-13i}{3}=\dfrac{15}{3}-\dfrac{13i}{3}\\\\\\Real \ part = \dfrac{15}{3} \ & \ imaginary \ part = \dfrac{-13}{3}

b) 1) 6 + 4i

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