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White raven [17]
3 years ago
13

Ieda bakes 73 cupcakes and accidentally drops 5 of them. She divides them equally into 13 containers for a bake sale for the sch

ool band. How many cupcakes does Ieda have left?
2 cupcakes
3 cupcakes
4 cupcakes
5 cupcakes
Mathematics
2 answers:
Norma-Jean [14]3 years ago
7 0
3 cupcakes:
When 5 drops, the remaining are;
73-5=68 cupcakes
13 containers will hold 65 cupcakes
Scrat [10]3 years ago
4 0
If Leda baked 73 cupcakes and accidentally drops 5. She has 68 left.

73 - 5 = 68

If she divides them in equal container she would have 3 left.

13 x 5 = 65
Therefore: 68/15= 5 r 3

Therefore Leda had 5 cupcakes in each container and 3 left.

Answer: 3 cupcakes
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If (2i/2+i)-(3i/3+i)=a+bi, then a=<br>A. 1/10<br>B. -10<br>C. 1/50<br>D. -1/10​
Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi

We need to find the value of a.

The LCM of (2+i) and (3+i)  is (2+i)(3+i)

=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}

Now rationalize the denominator by multiplying by 5-5i/5-5i

=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}

We are given

\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi

Now after solving we have:

\frac{1}{10}-\frac{i}{10}=a+bi

So value of a = 1/10 and value of b = -1/10

So, Option A is correct.

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