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2 years ago

HIGH SCHOOL GEOMETRY/ ALGEBRA

Mathematics

1 answer:

max2010maxim [7]2 years ago

3 0

Answer:

The ratio of the lateral sides and perimeters = the ratio of square root of their areas

Then, the ratio of the sides and the perimeters is 6:5 or if this is wrong it it is 5:6

if this helps plz mark as brainliest

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How many three-digit positive integers <img src="https://tex.z-dn.net/?f=x" id="TexFormula1" title="x" alt="x" align="absmiddle"

Anastaziya [24]

Answer:

Step-by-step explanation:

Any solution x will mod 23 will also have x+23n as a solution, for some integer n. Since 900/23 = 39 3/23, we know there are 39 or 40 three-digit integers of this form.

As it happens, 100 is the smallest 3-digit solution. So, there are 40 three-digit numbers that are of the form 100 +23n, hence 40 solutions to the equation.

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The equation reduces, mod 23, to ...

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Its solutions are x = 23n +8.

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2 years ago

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2 years ago

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Vikki [24]

Answer:

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Step-by-step explanation:

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3 years ago

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How do you convert repeating decimals into fractions

diamong [38]

Answer:So, To convert a decimal to a fraction, begin by placing that decimal over the number 1. Then keep multiplying both the numerator (top number) and denominator (bottom number) by 10 until both are whole numbers.

Step-by-step explanation:

So, To convert a decimal to a fraction, begin by placing that decimal over the number 1. Then keep multiplying both the numerator (top number) and denominator (bottom number) by 10 until both are whole numbers.

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3 years ago

Determine the zeroes of the polynomial

Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

$(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)} \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)} -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\$

$\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}$

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3 years ago