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Oksana_A [137]
3 years ago
11

Find the missing side of the triangle. Side LaTeX: cc is the hypotenuse and sides LaTeX: aa and LaTeX: bb are legs. Leave your a

nswer in simplest radical form. LaTeX: a=2,\:b=4
Mathematics
1 answer:
mylen [45]3 years ago
6 0

Answer:

c = 2\sqrt{5

Step-by-step explanation:

Given

a = 2

b = 4

c = hypotenuse

Required

Find c

Using Pythagoras theorem, we have:

c^2 = a^2 + b^2

Substitute values for a and b

c^2 = 2^2 + 4^2

c^2 = 4 + 16

c^2 = 20

Take square roots of both sides:

c = \sqrt{20

Express 20 as 4 * 5

c = \sqrt{4 * 5

Split surd

c = \sqrt{4} * \sqrt{5

Take square root of 4

c = 2 * \sqrt{5

c = 2\sqrt{5

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What is the remainder of (4x2 + 7x-1)= (4 + x)?<br>A. -9x – 1<br>B.23x – 1<br>C.35<br>D.-37​
bearhunter [10]

Answer: I don't know what you meant by remainder but i hope this helps :)

x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}\\

Step-by-step explanation:

\left(4x^2+7x-1\right)=\left(4+x\right)\\\mathrm{Refine}\\4x^2+7x-1=4+x\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\4x^2+7x-1-x=4+x-x\\Simplify\\4x^2+6x-1=4\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\4x^2+6x-1-4=4-4\\\mathrm{Simplify}\\4x^2+6x-5=0\\\mathrm{For\:}\quad a=4,\:b=6,\:c=-5:\\\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}

\frac{-6+\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\=\frac{-6+\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{8}\\\\Let\: simplify\: ; -6+2\sqrt{29}\\=-2\times \:3+2\sqrt{29}\\=2\left(-3+\sqrt{29}\right)\\=\frac{2\left(-3+\sqrt{29}\right)}{8}\\=\frac{-3+\sqrt{29}}{4}\\

\frac{-6-\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\\\=\frac{-6-\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\\\=\frac{-6-\sqrt{116}}{2\times \:4}\\\\=\frac{-6-2\sqrt{29}}{8}\\\\=-\frac{2\left(3+\sqrt{29}\right)}{8}\\\\=-\frac{3+\sqrt{29}}{4}\\\\\\x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}

6 0
3 years ago
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option c is correct answer friend

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