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Dennis_Churaev [7]
4 years ago
14

Richard’s checking account balance was $57.34 at the beginning of the week. During the week, he recorded the transactions below.

Mathematics
2 answers:
Tomtit [17]4 years ago
7 0

Answer:

US$ 132.45

Step-by-step explanation:

See attachment for the missing table.

Given:

Richard’s checking account balance at the beginning of the week = $57.34

<u>Richard’s account balance at the end of the week from the given table:</u>

Deposits of the week = US$ 163.75

Expenses of the week = Groceries + Credit card bill + Gas

Expenses of the week = 25.37 + 50 + 13.27

Expenses of the week = US$ 88.64

Richard’s account balance at the end of the week = Richard’s checking account balance at the beginning of the week + Deposits of the week - Expenses of the week

Replacing with the real values:

Richard’s account balance at the end of the week = 57.34 + 163.75 - 88.64

                                                                                    =US$ 132.4

GenaCL600 [577]4 years ago
4 0

nswer:132.45

Step-by-step explanation:

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Explanation:

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The denominator (x^2+4x-21)(x+1) updates to (x-3)(x+7)(x+1)

We have three pairs of terms that will cancel between the numerator and denominator. The (x+1) terms pair up and cancel out, so do the (x+7) terms and also the (x-3) terms as well. Everything cancels out. The only thing left is 1 because x/x = 1 where x is nonzero. We can have a more complicated expression replace x to get the same basic idea. This is what I mean by "canceling".

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What is the hypotenuse of a right triangle that has legs of 12x and 9x
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Step-by-step explanation:

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3 years ago
Factor completely 3x3 − 21x2 − 27x
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Answer: I dont understand this "27x", are you missing a number?

Hello!

______________________

3x3 − 21x2 − 27x

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4 years ago
A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccol
slamgirl [31]

Answer:

A. 6 188

B. 749 398

C. 6 188

D. 52 975

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F. 11 316

Explanation:

(a) The shop has 6 types of croissants of which a dozen(12) has to be selected

Therefore n=6, r=12

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188

(b) The shop has 6 types of croissants of which three dozen(36) has to be selected

Therefore n=6, r=36

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+36-1, 12) = C(41, 36) = 41!÷ 36!(41-36)! = 41!÷36! 5! = 749 398

(c) The shop has 6 types of croissants of which two dozen(24) has to be selected

Let us first select 2 of each kind which 12 croissants in total. Then we still need to select the remaining 12 croissants

Therefore n=6, r=12

Repetition of croissants is permitted

And C(n+r-1, r)

C(6+12-1, 12) = C(17, 12) = 17!÷ 12!(17-12)! = 17!÷12! 5! =6 188

(d) The shop has 5 types of croissants of which two dozen(24) has to be selected

Therefore n=5, r=24

Repetition of croissants is permitted

And C(n+r-1, r)

C(5+24-1, 24) = C(28, 24) = 28!÷ 24!(28-24)! = 28!÷24! 4! = 20 475

5 0
4 years ago
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