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hram777 [196]
3 years ago
11

Gabriel brings 25 cupcakes to share with his classmates at school. At the end of the day, he has 3 cupcakes remaining.

Mathematics
2 answers:
Gre4nikov [31]3 years ago
8 0

Answer:

d 88

Step-by-step explanation:

Mila [183]3 years ago
4 0

Answer:D

Step-by-step explanation:

88

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Find the value of the 60th term of the following sequence: 33, 28, 23, 18
Otrada [13]

Step-by-step explanation:

The first term of the AP i.e a = 33

and common difference i.e d = 28-33 = -5

So, 60th term = a +(n-1)d = a+ 59d

= 33 + 59( -5)

= 33- 295

= - 262

4 0
3 years ago
Find the zeros of the function h(x)=x^2-5x-50 by factoring
lukranit [14]
h(x)=x^2-5x-50=x^2-5x-25-25=x^2-25-5x-25\\\\=(x^2-25)-(5x+25)=(x^2-5^2)-5(x+5)\\\\=(x-5)\underline{(x+5)}-5\underline{(x+5)}=(x+5)(x-5-5)=(x+5)(x-10)\\\\Answer:Zeros\ are\ x=-5\ and\ x=10.
5 0
4 years ago
What is the inverse function of number 20 and how do you get the inverse?
mariarad [96]
I hope this helps you



(f (x))^2= (square root of x-2)^2



f^2 (x)=x-2


x=f^2 (x)+2


f^-1 (x)=x^2+2
7 0
3 years ago
At the end of October, Jane's credit card balance was $1,200. On November 16, she made $500 of purchases. On November 26, she pa
Elodia [21]

Answer:

Average day end balance using weighted average = $1,316.667

Average day end balance using simple average = $1,050

Step-by-step explanation:

Provided opening balance = $1,200

During the month addition in such balance = $500

New balance as on 16 November = $1,200 + $500 = $1,700

Further by the end of month of November balance after payment = $1,700 - $800

= $900

Thus, opening balance = $1,200

Closing = $900

Simple average = \frac{1,200 + 9,00}{2} = 1,050

Day end balance = $1,050

For this weighted average can be done

First 15 days balance = $1,200 as no transaction was done.

From 16  to 25 November balance increased by $500 = $1,700 for 10 days

Remaining days = 5 days = $1,700 - $800 paid = $900

Using weighted average we have

\frac{(1,200 \times 15) + (1,700 \times 10) + (900 \times 5)}{30} = \frac{39,500}{30} = 1,316.667

Day end balance = $1,316.667

3 0
3 years ago
A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not cons
Eduardwww [97]
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.

4 0
3 years ago
Read 2 more answers
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