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raketka [301]
3 years ago
15

Number 3 answered please

Mathematics
1 answer:
Darya [45]3 years ago
3 0

Answer:

Step-by-step explanation:

n=3p, d=p-2

10d+5n+p=110 using n and d from above makes this become

10(p-2)+5(3p)+p=110

10p-20+15p+p=110

26p-20=110

26p=130

p=5

n=3p=3(5)=15

d=p-2=5-2=3

So she has 3 dimes, 15 nickels, and 5 pennies.

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muminat

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1st square =0

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3 years ago
I need help! Whoever answers get brainlist. Please write a step by step so I know how to do it next time.
alexgriva [62]

Answer: 8/3

Step-by-step explanation:

So first what you do is you have to add the whole number to the numerator and you do that by multiplying the whole number by the deniminator which is 6. Now you add the 6 to the numerator which is 8, so you get 8/3.

Sorry this got so complicated. Hope this helped!

 

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Data were collected every five years on the population of the countries in the UN. The following computer output and residual pl
Vladimir [108]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
Name two coordinates that are on the line: <img src="https://tex.z-dn.net/?f=y%20%3D%2012" id="TexFormula1" title="y = 12" alt="
Eva8 [605]

Answer:

(1,12) and (2,12).

(- 5,1) and (- 5, 2).

Step-by-step explanation:

y = 12 is a line that is parallel to the x-axis and at a constant distance of 12 units above the x-axis. Therefore, the points on the line will have any x-coordinate but have a constant y-coordinate i.e. 12.

Therefore, the two points on the line y = 12 can be (1,12) and (2,12).  

x = - 5 is a line that is parallel to the y-axis and at a constant distance of 5 units left of the y-axis. Therefore, the points on the line will have any y-coordinate but have a constant x-coordinate i.e. - 5.

Therefore, the two points on the line x = - 5 can be (- 5,1) and (- 5, 2).  (Answer)

3 0
3 years ago
5x-4y-6z = -3<br> X-3y+z=-1<br> -3x-6y+7z=1
PilotLPTM [1.2K]

Answer:

x=-\frac{41}{55},y=\frac{2}{55},z=-\frac{8}{55}

Step-by-step explanation:

5x-4y-6z=-3...............eq(1)\\\\x-3y+z=-1.....................eq(2)\\\\-3x-6y+7z=1.................eq(3)

eq(1)-5\times eq(2)

5x-4y-6z-5(x-3y+z)=-3-5\times (-1)\\\\11y-11z=2\\\\y-z=\frac{2}{11}...................eq(4)

3\times eq(2)+eq(3)

3(x-3y+z)-3x-6y+7z=3\times (-1)+1\\\\-15y+10z=-2\\\\-y+\frac{2}{3}z=-\frac{2}{15}.........eq(5)

eq(4)+eq(5)

y-z-y+\frac{2}{3}z=\frac{2}{11}-\frac{2}{15}\\\\-\frac{1}{3}z=\frac{8}{165}\\\\z=-\frac{8\times 3}{165}\\\\z=-\frac{8}{55}

From eq(4)

y=z+\frac{2}{11}=-\frac{8}{55}+\frac{2}{11}\\\\y=\frac{2}{55}

Substitute the value of y and z in eq(2)

x-3\times \frac{2}{55}-\frac{8}{55}=-1\\\\x-\frac{14}{55}=-1\\\\x=-1+\frac{14}{55}\\\\x=\frac{41}{55}

6 0
4 years ago
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