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jenyasd209 [6]
3 years ago
5

What numbers are divisible by 2 and 5 but not 3

Mathematics
1 answer:
serious [3.7K]3 years ago
7 0

Answer:

10  20 40

Step-by-step explanation:

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Solve for x.<br><br> 5.2(x−6.7)=46.28
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5.2(x - 6.7) = 46.28

Divide both sides of the equation by 5.2

x - 6.7 = 8.9

Move the constant to the right-hand side and change its sign

x = 8.9 + 6.7

Add the decimals

x = 15.6

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Write this radical in simplest form:
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Answer:

in simplest form the radical would be, 4 3/4

( the / is supposed to be the square root sign)

hope this helped ;D

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Which of the following is equivalent to (x-3)(x^2-2x-3)
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What is the following
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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}
4 0
3 years ago
Help please help please​
Colt1911 [192]

Answer:

+7 or 7

Step-by-step explanation:

an integer is a whole number. So the question asks what whole number represent this real world situation. A positive charge to an atom can not be represent by a negative number (number below zero). Therefore, it has to be positive.

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