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Yuliya22 [10]
3 years ago
10

The different kinds of beads Jen is using to make purses come in packages of 4 and 12. What is the least number of each kind of

bead Jen can buy to have an equal number of each of the different kinds of​ beads?
Mathematics
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

You should buy 3 packages of beads that come in packages of 4 and 1 package of beads that come in packages of 12 to have an equal number of each of the different kinds of beads.

Step-by-step explanation:

Since the different kinds of beads Jen is using to make purses come in packages of 4 and 12, to determine what is the least number of each kind of bead Jen can buy to have an equal number of each of the different kinds of beads the following calculation must be performed:

12/4 = 3

4 x 3 = 12

Thus, you should buy 3 packages of beads that come in packages of 4 and 1 package of beads that come in packages of 12 to have an equal number of each of the different kinds of beads.

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94 bagels in total. 100 will be plenty

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What value will make the equation true -21-?=1.5
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Answer: ? = -22.5

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6 0
3 years ago
Yes :]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]
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It would be 150 I believe.

Step-by-step explanation:

Hope this helps and good luck!

6 0
3 years ago
Rationalize the denominator and simplify.
cestrela7 [59]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3151018

——————————
 
     \mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}


Multiply and divide by the conjugate of the denominator, which is  \mathsf{4\sqrt{6}+3\sqrt{2}:}

     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}


Expand those products and eliminate the brackets:
 
     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}

     \mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}} 

     \mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}} 

     \mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

4 0
3 years ago
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