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valentinak56 [21]
2 years ago
8

How do know if you can divide a number by 3. Is there a pattern?

Mathematics
2 answers:
ioda2 years ago
7 0
When you add up the sum of the digits, it should be a multiple of three. 21 is 2 plus 1 equals 3, so since 3 is a multiple of three, 21 is divisible by 3
kvv77 [185]2 years ago
4 0
You can know if the number is divisible by 3 by adding the numbers up. If the sum is dividable by 3 then the original number can be as well. For example, the number 93. 9+3= 12 and 12 can be divided by 3. This means 93 is also divisble by three. If you do divide 93 the quotient is 31.
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Find the radius of the circle whose equation is (x + 5)2 + (y + 8)2 = 16

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Which one of the following would be a good tool to represent data in terms of percentage of a whole?
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I think the correct answer from the choices listed above is the first option. It is a pie chart that would be <span>a good tool to represent data in terms of percentage of a whole. It is a graph that makes use of a circle divided into parts that represents the proportion of the whole.</span>
6 0
2 years ago
Read 2 more answers
Multiply.
Crank
The main rules that we use here are :

i) \sqrt{a \cdot b}=  \sqrt{a} \cdot \sqrt{b} for nonnegative values a and b.

ii) \sqrt{a} \cdot  \sqrt{a} =a.


Thus, first 'decompose' the numbers in the radicals into prime factors:

4 \sqrt{3} \cdot 10 \cdot  \sqrt{2\cdot2\cdot3}\cdot  \sqrt{2\cdot 3}\cdot  \sqrt{2}..

By rule (i) we write:

4 \sqrt{3} \cdot 10 \cdot \sqrt{2}\cdot \sqrt{2}\cdot \sqrt{\cdot3}\cdot \sqrt{2} \cdot  \sqrt{3} \cdot \sqrt{2}.

We can collect these terms as follows:

40 \sqrt{3}\cdot (\sqrt{3}\cdot \sqrt{3}) \cdot (\sqrt{2}\cdot \sqrt{2}) \cdot( \sqrt{2} \cdot \sqrt{2}), and by rule (ii) we have:

40 \sqrt{3}\cdot 3 \cdot 2 \cdot2=40\cdot12\cdot \sqrt{3}=480 \sqrt{3}.


Answer: 480 \sqrt{3}.
8 0
3 years ago
Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!
il63 [147K]

Answer:

x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

Step-by-step explanation:

w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

6 0
2 years ago
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