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Genrish500 [490]
3 years ago
14

Ski club members are preparing identical welcome kits for new skiers. the ski club has 60 hand-warmer packets and 48 foot-warmer

packets. find the greatest number of identical kits they can prepare using all of hand warmer and foot warmer packets. how many hand warmer packets and foot warmer packets would each welcome kit have
Mathematics
1 answer:
kvasek [131]3 years ago
8 0
You have to find the greatest common factor GCF for each number, you can do it for both at the same time:

             foot ----  hand
GCF       48          60
   2          24          30
   2          12          15
   3           4            5
From here you dont have more numbers (GCF) for this pair, now for finding the greatest number of identical kits multiply the GCFs: 2*2*3=12 you have 12 kits containing 4 foot-warmer packets and 5 hand-warmer packets (from table the last row in foot and hand)
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Answer:

24

Step-by-step explanation:

6+6 =12

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18+6 = 24

7 0
3 years ago
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what is the vertical height of interverted cone with the diameter of 6 cm and the slant height of 6cm?
den301095 [7]

"interverted" or "inverted?"

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Then h^2 + (3 cm)^2 = (6 cm)^2, or h^2 + 9 cm^2 = 36 cm^2, or h^2 = 27 cm^2.

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3 years ago
How do I do 8b(ii) ? Please help me thank you!
Mila [183]
Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2      Multiply through by 4
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OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
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h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


7 0
3 years ago
Consider the two expressions 4b(b+1) and (2b+7)(2b-8). Compare their values if b=-3, b=-2, and if b=10. Is it true that for an v
frutty [35]

Answer:

FIRST EXPRESSION:

-  If b=-3, the value of 4b(b+1) is  24

-  If b=-2, the value of 4b(b+1) is  8

- If b=10, the value of 4b(b+1) is  440

 SECOND EXPRESSION:

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Yes, for any value of "b" the value of the first expression is greater than the value of the second expression.

Step-by-step explanation:

Substitute the given values of "b" into each expression and evaluate.

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If b=-3 → (2(-3)+7)(2(-3)-8)=-14

If b=-2 → (2(-2)+7)(2(-2)-8)=-36

 If b=10 → (2(10)+7)(2(10)-8)=324

You can observe that for any value of "b" the value of the first expression is greater than the value of the second expression.

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