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Lemur [1.5K]
3 years ago
10

What is the answer for -2y=3/4

Mathematics
1 answer:
ElenaW [278]3 years ago
8 0

Answer: y= -3/8

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation

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If a and B are the zeros of the quadratic polynomial f(x) = x2- 5x + 4 find the value of 1/a+1/b-2ab
bearhunter [10]

Answer:

<h2>-27/4</h2>

Step-by-step explanation:

Given the quadratic polynomial given as g(x) = x²- 5x + 4, the zeros of the quadratic polynomial occurs at g(x) = 0 such that x²- 5x + 4 = 0.

Factorizing the resulting equation to get the roots

x²- 5x + 4 = 0

(x²- x)-(4 x + 4) = 0

x(x-1)-4(x-1) = 0

(x-1)(x-4) = 0

x-1 = 0 and x-4 = 0

x = 1 and x = 4

Since a and b are known to be the root then we can say a = 1 and b =4

Substituting the given values into the equation  1/a+1/b-2 ab , we will have;

= 1/1 + 1/4 - 2*1*4

= 1 + 1/4 - 8

= 5/4 - 8

Find the Lowest common multiple

= (5-32)/4

= -27/4

<em>Hence the required value is -27/4</em>

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3 years ago
Mary is writing invitations for her birthday party. She plans to invite 16 friends. So far, Mary has written 12 invitations. Whi
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The answer is 75% and 75/100

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Answer:

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Step-by-step explanation:

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A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
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