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Julli [10]
4 years ago
6

What’s the answer for (3b+3c) 2 for b=4 and c=5

Mathematics
2 answers:
vivado [14]4 years ago
6 0

Answer:

The answer is:54

Step-by-step explanation:

U first multiply 3*4 then u multiply  3*5 then add them together then multiply them by 2


Airida [17]4 years ago
3 0

Answer:

54

Step-by-step explanation:

substitute 4 for b and 5 for c and you will get ( 3(4) + 3(5) ) 2.......3 times 4 is 12 and 3 times 5 is 15 and 12 + 15 is 27 and 27 times 2 is 54

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-X/10 = 1/5<br> What is the answer
Georgia [21]
The answer is x=-2

Solve for X by simplifying both sides of the equation, then isolating the variable.
7 0
3 years ago
Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6
kenny6666 [7]

Answer:

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

Step-by-step explanation:

Given the simultaneous equations

y=9-x

y\:=\:2x^2\:+\:4x\:+\:6

Subtract the equations

y=9-x

-

\underline{y=2x^2+4x+6}

y-y=9-x-\left(2x^2+4x+6\right)

\mathrm{Refine}

x\left(2x+5\right)=3

\mathrm{Solve\:}\:x\left(2x+5\right)=3

2x^2+5x=3        ∵ \mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x

\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

2x^2+5x-3=3-3

\mathrm{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=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\

x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

  =\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{4}

  =\frac{-5+7}{4}

  =\frac{2}{4}

  =\frac{1}{2}

Similarly,

x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{1}{2},\:x=-3

\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

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Answer:

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Step-by-step explanation:

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da first one: .1

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7 0
3 years ago
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A bag has 6 yellow marbles and 6 brown marbles. Half of the brown marbles are made of glass. A marble is selected at random from
AysviL [449]
Total = 12 marbles
6 brown marbles but h<span>alf of the brown marbles are made of glass
so brown and made of glass = 3
</span><span>probability that it is a brown, glass marble = 3/12 = 1/4

answer
1/4</span>
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