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luda_lava [24]
3 years ago
6

Juanita is making necklaces to give as presents. She plans to put 15 beads on each necklace. Beadsare sold in packages of 20. Wh

at is the least number of packages she can tp make necklaces and have no beads left over?
Mathematics
2 answers:
PtichkaEL [24]3 years ago
4 0
3 packages because you'll have 15 beads left over. which would complete the use of the packages
Alex Ar [27]3 years ago
3 0
3 packages is the anwer hopefully this helped
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Can anyone figure this out?
Verizon [17]

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill


\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}


now that we know how long each one is, let's plug those in Heron's Area formula.


\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18

5 0
3 years ago
X-3y=-24<br> 5x+8y=-5<br><br> this equation is substitution, need help :))
Leto [7]

Answer:

x=-9, y=5. (-9, 5).

Step-by-step explanation:

x-3y=-24

5x+8y=-5

--------------

x=-24+3y

5(-24+3y)+8y=-5

-120+15y+8y=-5

-120+23y=-5

23y=-5-(-120)

23y=-5+120

23y=115

y=115/23

y=5

x-3(5)=-24

x-15=-24

x=-24+15

x=-9

6 0
3 years ago
6m + 2 &lt; 5m-4 solve the inequality for m
djyliett [7]

Answer:

m < -6

Step-by-step explanation:

plz can u mark brainliest if correct. thank you so much.

6 0
3 years ago
A catering business offers two sizes of baked ziti. Its small ziti dish uses 1 cup of sauce
castortr0y [4]

Answer:

x + 2y ≤ 100 and x + 3y ≤ 400

Maximum profit = 6x + 5y.

Step-by-step explanation:

Let there be x number of small dishes and y number of large dishes to maximize the profit.

So, total profit is P = 6x + 5y .......... (1)

Now, the small dish uses 1 cup of sauce and 1 cup of cheese and the large dish uses 2 cups of sauce and 3 cups of cheese.

So, as per given conditions,  

x + 2y ≤ 100 ........ (1) and  

x + 3y ≤ 400 .......... (2)

Therefore, those are the constraints for the problem. (Answer)

4 0
2 years ago
Why couldn't you add the<br> feet, yards, and centimeters<br> together?
I am Lyosha [343]
It’s not possible because it needs to be convicted to feet then yards then centimeter but you can add it you will get something way different I tried before
8 0
2 years ago
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