<span>23 identical packets, consisting of 1 sheet of paper and 4 crayons.
The maximum number of packets the teacher can make from 92 crayons and 23 sheets of paper is the greatest common factor of 92 and 23. And for the values of 92 and 23, the GCF is 23.
First, calculate the list of prime numbers that when multiplied together, get each number.
23 = 23
92 = 2 * 2 * 23
Look for the largest set of prime factors in common to both numbers. In this case, it's 23.
So the teach can make 23 identical activity packets. Each packet will contain (92 / 23 = 4) crayons, and (23 / 23 = 1) sheet of paper.</span>
Answer:
<u>f</u><u>(</u><u>-</u><u>2</u><u>)</u><u> </u><u>+</u><u> </u><u>9</u><u> </u><u>is</u><u> </u><u>1</u><u>9</u>
Step-by-step explanation:
![f(x) = - 3x - 2](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20-%203x%20-%202)
when x is -9:
![f( - 2) = - 3( - 2) - 2 \\ f( - 2) = 10](https://tex.z-dn.net/?f=f%28%20-%202%29%20%3D%20%20-%203%28%20-%202%29%20-%202%20%5C%5C%20f%28%20-%202%29%20%3D%2010)
sum up with 9, f(-2)+9:
![f( - 2) + 9 = 10 + 9 \\ = 19](https://tex.z-dn.net/?f=f%28%20-%202%29%20%2B%209%20%3D%2010%20%2B%209%20%5C%5C%20%20%3D%2019)
Is there an image? Or a multiple choice?
Answer:
Lenght = 13, Width = 4
Step-by-step explanation:
Perimeter:
2(L + W) = 34
=> 2L + 2W = 34
Area:
L * W = 52
L = 2W + 5
L - 2W = 5
----------------------------
How to find the Length:
Now Add:
2L + 2W = 34
+ L - 2W = 5
-------------------------
3L = 39
L = 13
How to find the Width:
L * W = 52
13 * W = 52
W = 4
======================
Length = 13
Width = 4