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Kitty [74]
2 years ago
9

One ingot contains 10 kg of pure silver and 2 kg of league. What quantity of silver, whose grade is 0.700; is it necessary to me

lt to obtain silver with a grade of 0.750?
a) 20
b) 22
c) 16
d) 24
e) 19
Mathematics
1 answer:
gulaghasi [49]2 years ago
4 0

Answer:

a) 20

Step-by-step explanation:

If x is the kg of 0.700 grade silver, then:

Silver in ingot + silver in 0.700 alloy = silver in 0.750 alloy

10 + 0.7x = 0.75(x + 12)

10 + 0.7x = 0.75x + 9

1 = 0.05x

x = 20

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Lubov Fominskaja [6]
Let the smaller number be x
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x²=36
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x = 6 

x=6
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5x = 30

So the two numbers are 6 and 30
7 0
2 years ago
A rectangular prism has a length of 9 feet, a height of 20 feet, and a width of 4 feet. What is its volume, in cubic feet?
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Answer:

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Step-by-step explanation:

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happy to help

7 0
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irina [24]

Answer:

Step-by-step explanation:

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Complete the table of values
Agata [3.3K]
Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify! 

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, 3^{-2} =&#10;\frac{1}{3^{2} }.
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, (\frac{3}{4}) ^{3}  =  \frac{ 3^{3} }{4^{3} }.
3) The zero exponent rule<span> says that any number raised to zero is 1. For example, 3^{0} = 1.
</span>

Back to the Problem:
Problem 1 
The x-values are in the left column. The title of the right column tells you that the function is y =  4^{-x}. The x-values are:
<span>1) x = 0
</span>Plug this into y = 4^{-x} to find letter a:
y = 4^{-x}\\&#10;y = 4^{-0}\\&#10;y = 4^{0}\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = 4^{-x} to find letter b:
y = 4^{-x}\\ &#10;y = 4^{-2}\\ &#10;y =  \frac{1}{4^{2}} \\  &#10;y= \frac{1}{16}
<span>
3) x = 4
</span>Plug this into y = 4^{-x} to find letter c:
y = 4^{-x}\\ &#10;y = 4^{-4}\\ &#10;y =  \frac{1}{4^{4}} \\  &#10;y= \frac{1}{256}
<span>

Problem 2
</span>The x-values are in the left column. The title of the right column tells you that the function is y =  (\frac{2}{3})^x. The x-values are:
<span>1) x = 0
</span>Plug this into y = (\frac{2}{3})^x to find letter d:
y = (\frac{2}{3})^x\\&#10;y = (\frac{2}{3})^0\\&#10;y = 1
<span>
2) x = 2
</span>Plug this into y = (\frac{2}{3})^x to find letter e:
y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\&#10;y =  \frac{4}{9}
<span>
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y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}
<span>
-------

Answers: 
a = 1
b = </span>\frac{1}{16}<span>
c = </span>\frac{1}{256}
d = 1
e = \frac{4}{9}
f = \frac{16}{81}
5 0
3 years ago
100 points!!!
Bogdan [553]

Answer:

(1, 5)

Step-by-step explanation:

The solution to the system of equations is the point of intersection of the two lines.  From inspection of the graph, the point of intersection is at (1, 5).

<u>Proof</u>

The solution to a system of equations is the point at which the two lines meet.  

⇒ g(x) = f(x)

⇒ 3x + 2 = |x - 4| + 2

⇒ 3x = |x - 4|

⇒ 3x = x - 4   and   3x = -(x - 4)

⇒ 3x = x - 4

⇒ 2x = -4

⇒ x = -2

Inputting x = -2 into the 2 equations:

⇒ g(-2) = 3 · -2 + 2 = -4

⇒ f(-2) = |-2 - 4| + 2 = 8

Therefore, as the y-values are different, x = -2 is NOT a solution

⇒ 3x = -(x - 4)

⇒ 3x = 4 - x

⇒ 4x = 4

⇒ x = 1

Inputting x = 1 into the 2 equations:

⇒ g(1) = 3 · 1 + 2 = 5

⇒ f(1) = |1 - 4| + 2 = 5

Therefore, as the y-values are the same, x = 1  IS a solution

and the solution is (1, 5)

8 0
2 years ago
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