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larisa [96]
3 years ago
5

Sami purchased 5 comic books priced as shown.

Mathematics
2 answers:
Vedmedyk [2.9K]3 years ago
8 0
19.21

adding all the totals together gives you 17.63 . the sales tax is 1.5867 . add that to the 17.63 and you’ll get 19.21
kati45 [8]3 years ago
6 0

Answer:

$19.23

Step-by-step explanation:

First, add the prices of all 5 comics together. That should equal $17.63. The tax is 9% so that would be 1.5867 or $1.60. In total, Sami paid $19.23.

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I need this please help
Y_Kistochka [10]
Function A has a rate of change of 0.75 while Function B has a rate of change of 0.25 so Function A has a greater rate of change.
5 0
3 years ago
If Emma is 12 Luke is two times her age Kendra is 1 half of Luke's age how old is Kendra
Natali [406]
We already know that Emma is 12, and if Luke is twice her age, then that means we need to multiply 12 by 2, which equals 24. If Kendra is one half of Luke's age, which is 24, then we must divide 24 by 2, which equals 12.
In conclusion, Kendra is 12 years old.
5 0
3 years ago
Read 2 more answers
a recipe calls for 6 cups of water and 4 cups of flour. if the recipe is decreased to use 2 cups of water, how much flour should
Len [333]
\frac{6}{4}=\frac{2}{f}\\\frac{3}{2}=\frac{2*\frac{3}{2}}{f*\frac{3}{2}}\\\frac{3}{2}=\frac{3}{\frac{3}{2}f}\\2=\frac{3}{2}f\\2*\frac{2}{3}=\frac{3}{2}f*\frac{2}{3}\\\frac{4}{3}=f

1 1/3 cups of flour
3 0
3 years ago
Help please!!!!
Mrac [35]

Answer:

a = 2

b = 18

a/b = 1/9

Step-by-step explanation:

5 0
3 years ago
What is the sum of the first 70 consecutive odd numbers? Explain.
expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N

So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

5 0
3 years ago
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