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disa [49]
3 years ago
8

the list shows how much yarn mary has= 1/3 red yarn, 2/9 white yarn, 2/6 blue yarn what is the total amount of yarn mary has?

Mathematics
1 answer:
zlopas [31]3 years ago
4 0

The total amount of yarn Mary has is \frac{8}{9}.

Step-by-step explanation:

Mary has;

Red yarn = \frac{1}{3}

White yarn = \frac{2}{9}

Blue yarn = \frac{2}{6}

Total yarn = Red + White + Blue

Total yarn = \frac{1}{3}+\frac{2}{9}+\frac{2}{6}\\

Taking LCM = 18

Total yarn = \frac{6+4+6}{18}

Total yarn = \frac{16}{18}=\frac{8}{9}

The total amount of yarn Mary has is \frac{8}{9}.

Keywords: fraction, addition

Learn more about fractions at:

  • brainly.com/question/2150928
  • brainly.com/question/2154850

#LearnwithBrainly

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Which could be the resulting equation when elimination is used to solve the given system of equations ?
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Answer:

10·b = 60

Step-by-step explanation:

The given system of equation is presented as follows;

5·a + 5·b = 25...(1)

-5·a + 5·b = 35...(2)

Given that the coefficient of a in equation (1) is equal in magnitude but opposite in sign to the coefficient of 'a' in equation (2), to eliminate the variable 'a' when using the elimination method, we add both equations as follows;

5·a + 5·b + (-5·a + 5·b) = 25 + 35 = 60

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5 0
3 years ago
What is the square root of -2i?
Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of z=0-2i. First, convert from Cartesian to polar form:

r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2

\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}

z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})

Next, use the formula \displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr] where \displaystyle k=0,1,2,...\:,n-1 to find the square roots:

<u>When k=1</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]

\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)

\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i

<u>When k=0</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)

\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i

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4 0
2 years ago
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2 years ago
Given that g(x)=3x²-5x+7 find the following: <br> g(-x)
AURORKA [14]
G(x) = 3x² - 5x + 7
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This equation cannot be solved because of a few reasons,

1. This equation didn't show that it equals to 0.
2. Even if it equals to zero, square root of a negative number cannot be solved.
(I will show you what I mean)

-3x² + 5x - 7
is the same as
3x² - 5x + 7
by shifting the equation,

for example,
1 - 3 = -2
shifting other side
2 = -1 + 3

using 3x² - 5x + 7 to solve,

Solve\ using\ the\ formular\ \ \boxed{ x= { \frac{-b \pm \sqrt{b^2-4ac} }{2a} } }
a = 3
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x= { \frac {-(-5) \pm \sqrt{ (-5)^2-4(3)(7) } }{2(3)} }
x= { \frac {5 \pm \sqrt{-59} }{6} }

∴This equation cannot be solved.
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