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mrs_skeptik [129]
3 years ago
15

Frederick is working on a number puzzle and discovers that the product of the two base numbers is exactly twice as large as the

sum of those same base numbers. Help on this, please?

Mathematics
2 answers:
Effectus [21]3 years ago
7 0
X*y=2(x+y)

This is an equation representing your word problem. X and Y are the two base numbers, assuming they are different.
poizon [28]3 years ago
7 0

Let two base numbers that is any of the digit from [0-9] be x and y.

  It is given that,  the product of the two base numbers is exactly twice as large as the sum of those same base numbers.

→xy=2×(x+y)

There are two variables in the equation , and there is only one equation.So, we have to use Hit and Trial method to evaluate the value of x and y.

 → 0≤ x≤9 and 0≤ y≤9

⇒ Integral Solution of the above equation can be evaluated by drawing the graph  (Drawn the graph using Desmos)

1.⇒ x=4 and , y=4.

2.⇒ x=6 and y=3

3.⇒x=3 and y=6

4.⇒x=0 and y=0

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igor_vitrenko [27]

Answer:

She can make three shirts and will have 7/56 left.

Step-by-step explanation:

49/56-14/56=35/56 (one shirt) 35/56-14/56=21/56 (two shirts) 21/56-14/56=7/56 (three shirts)

8 0
3 years ago
A music store bought a CD set at a cost of​ $20. When the store sold the CD​ set, the percent markup was​ 40%. Find the selling
Debora [2.8K]

Answer:

$28

Step-by-step explanation:

cp=20

mp=20%ofcp

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8 0
2 years ago
What is the solution to the equation x+473 = 0
nikitadnepr [17]
-473 + 473=0 , so D.
5 0
2 years ago
Read 2 more answers
PLEASE HELP (this is all my points I’ll mark brainlist anything please help)
timurjin [86]

Answer:

I believe that it would be A+C+S but is there more to the question?

Step-by-step explanation:

5 0
3 years ago
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If a+b+c=0 then find the value of a^2+b^2+c^2/a^2-bc pls help me
VladimirAG [237]

a+b+c=0

[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]

[a^2+b^2+c^2+2ab+2ac+2bc=0]

[a^2+b^2+c^2=-(2ab+2ac+2bc)]

[a^2+b^2+c^2=-2(ab+ac+bc)] (i)

also

[a=-b-c]

[a^2=-ab-ac] (ii)

[-c=a+b]

[-bc=ab+b^2] (iii)

adding (ii) and (iii) ,we have

[a^2-bc=b^2-ac] (iv)

devide (i) by (iv)

[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
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3 years ago
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