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Sliva [168]
3 years ago
12

A: 105 B : 1,050 C: 9,450 D: 21

Mathematics
2 answers:
ad-work [718]3 years ago
7 0
The answer is B, 1,050 :)
charle [14.2K]3 years ago
4 0

Answer:

1050

Step-by-step explanation:

We can use a  ratio to solve

First find how many jam

210 - 189 =21

We want to find how many jam for 10500 printed

21 jam              x jam

-----------       = ----------------

210 printed          10500 printed

Using cross products

21 * 10500 = 210x

Divide each side by 210

21*10500/210 =x

1050=x

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2ᵃ = 5ᵇ = 10ⁿ.<br> Show that n = <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bab%7D%7Ba%20%2B%20b%7D%20" id="TexFormula1" titl
11Alexandr11 [23.1K]
There are two ways you can go about this: I'll explain both ways.
<span>
</span><span>Solution 1: Using logarithmic properties
</span>The first way is to use logarithmic properties.

We can take the natural logarithm to all three terms to utilise our exponents.

Hence, ln2ᵃ = ln5ᵇ = ln10ⁿ becomes:
aln2 = bln5 = nln10.

What's so neat about ln10 is that it's ln(5·2).
Using our logarithmic rule (log(ab) = log(a) + log(b),
we can rewrite it as aln2 = bln5 = n(ln2 + ln5)

Since it's equal (given to us), we can let it all equal to another variable "c".

So, c = aln2 = bln5 = n(ln2 + ln5) and the reason why we do this, is so that we may find ln2 and ln5 respectively.

c = aln2; ln2 = \frac{c}{a}
c = bln5; ln5 = \frac{c}{b}

Hence, c = n(ln2 + ln5) = n(\frac{c}{a} + \frac{c}{b})
Factorise c outside on the right hand side.

c = cn(\frac{1}{a} + \frac{1}{b})
1 = n(\frac{1}{a} + \frac{1}{b})
\frac{1}{n} = \frac{1}{a} + \frac{1}{b}

\frac{1}{n} = \frac{a + b}{ab}
and thus, n = \frac{ab}{a + b}

<span>Solution 2: Using exponent rules
</span>In this solution, we'll be taking advantage of exponents.

So, let c = 2ᵃ = 5ᵇ = 10ⁿ
Since c = 2ᵃ, 2 = \sqrt[a]{c} = c^{\frac{1}{a}}

Then, 5 = c^{\frac{1}{b}}
and 10 = c^{\frac{1}{n}}

But, 10 = 5·2, so 10 = c^{\frac{1}{b}}·c^{\frac{1}{a}}
∴ c^{\frac{1}{n}} = c^{\frac{1}{b}}·c^{\frac{1}{a}}

\frac{1}{n} = \frac{1}{a} + \frac{1}{b}
and n = \frac{ab}{a + b}
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