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Dovator [93]
2 years ago
7

How much of an alloy that is 30% copper should be mixed with 50 ounces of an alloy that is 60% copper in order to get an alloy t

hat is 40% ​copper?
Mathematics
2 answers:
kompoz [17]2 years ago
4 0

Answer: 180 i think

Step-by-step explanation: .5*200+.2X=.3(200+X)

100+.2X=60+.3X

.2X-.3X=60-100

-.1X=-40

X=-40/-.1

X=400 OUNCES OF 20% COPPER.

PROOF

.5*200+.2*400=.3(200+400)

100+80=.3*600

100+80=180

180=180

erik [133]2 years ago
4 0

Answer: 100 ounces of the alloy.

Step-by-step explanation: To solve, we create an equation:

First, though, let's define some variables and values -

x = amount of the alloy we are trying to find

0.3 = 30% copper

0.6 = 60% copper

50 = 50 ounces of the alloy

0.3(x) + 0.6(50) = 0.4(x+50) - original equation

0.3x + 0.6*50 = 0.4x + 0.4*50 - expanded

subtract 0.3x and 0.4*50 on both sides to get...

0.2*50 = 0.1x - simplified

calculate...

10 = 0.1x - simplified

divide both sides by 0.1 to get...

100 = x --> x = 100

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200 3. A technician charges $25 per hour plus $50 for a house call to repair home computers. Make a table and a graph to show th
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6 0
1 year ago
Choose ALL possible answers
muminat
Select all the correct answers:
1) Yes

2) No
x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170
x=8→f(8)=3^8+2=6,561+2→f(8)=5,563>170=h(8)

3) Yes

4) No

5) Yes
rg=[g(3)-g(2)]/(3-2)=[g(3)-g(2)]/1→rg=g(3)-g(2)
g(3)=20(3)+4=60+4→g(3)=64
g(2)=20(2)+4=40+4→g(2)=44
rg=64-44→rg=20

rf=f(3)-f(2)
f(3)=3^3+2=27+2→f(3)=29
f(2)=3^2+2=9+2→f(2)=11
rf=29-11→rf=18

rh=h(3)-h(2)
h(3)=2(3)^2+5(3)+2=2(9)+15+2=18+15+2→h(3)=35
h(2)=2(2)^2+5(2)+2=2(4)+10+2=8+10+2→h(2)=20
rh=35-20→rh=15

rg=20>18=rf
rg=20>15=rh

6)  No
x=4→g(4)=20(4)+4=80+4→g(4)=84
x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54
x=4→f(4)=3^4+2=81+2→f(4)=83>54=h(4)
f(4)=83<84=g(4)
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Step-by-step explanation:

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