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Ludmilka [50]
3 years ago
10

An alloy consists of nickel, zinc, and copper in the ratio 2:7:9. How many pounds of nickel have to be used to create alloy that

contains 4.9 lb of zinc?
Mathematics
2 answers:
Cloud [144]3 years ago
4 0

ANSWER

Find out the  how many pounds of nickel have to be used to create alloy that contains 4.9 lb of zinc.

To proof

As given

An alloy consists of nickel, zinc, and copper in the ratio 2:7:9.

the total part of the alloy are 18.

assuming one part be x.

thus

nickel part = 2x

zinc part = 7x

copper part = 9x

alloy that contains 4.9 lb of zinc.

than the equation becomes

7x = 4.9

x = \frac{4.9}{7}

x = 0.7

Therefore

nickel part = 2x

                  = 2× 0.7

                  = 1.4 lb

1.4 lb of nickel have to be used to create alloy that contains 4.9 lb of zinc.

Hence proved

KonstantinChe [14]3 years ago
3 0

Answer:

1.08 lb

Step-by-step explanation:

let the quantity of nickel, zinc and copper in the alloy be 2x, 7x and 9x respectively.


as per the question, amount of zinc in the alloy is 4.9 lb

so, 9x = 4.9

x = 4.9 / 9

x = 0.54


so, quantity of nickel = 2x = 2 *.54 = 1.08

so, pounds of nickel required to create alloy = 1.08 lb

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What is the answer to this?
amm1812

Answer:

<h2>40.1 cm</h2>

solution,

BC=8.4 cm

AB=12 cm

BD=4.1 cm

We have to find out: AC ,CD

using Pythagoras theorem:

In ∆ABC ,<B=90°

{(ac)}^{2}  =  {(ab)}^{2}  +  {(bc)}^{2} \\  {(ac)}^{2}   =  {(12)}^{2}  +  {(8.4)}^{2}  \\  {(ac)}^{2}  = 144 + 70.56 \\  {(ac)}^{2}  = 214.56 \\ ac =  \sqrt{214.56}  \\ ac = 14.64 \: cm

Similarly,

In ∆BCD, CB=90

cd =   \sqrt{ {(8.4)}^{2} +  {(4.1)}^{2}  }

=  \sqrt{70.56 + 16.81}  \\  =  \sqrt{87.37}  \\  = 9.34 \: cm

Perimeter:

AC+CD+AD

=14.64+9.34+(12+4.1)

=40.08

= 40.1 cm

Hope this helps...

Good luck on your assignment..

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