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garri49 [273]
3 years ago
5

1) work out the value of (1.7x10^4)x (8.5x10^-2) give your answer in standard form.

Mathematics
1 answer:
Mashutka [201]3 years ago
8 0

ANSWER TO QUESTION 1

(1.7\times 10^4)\times(8.5\times 10^{-2})


We rewrite to obtain;


(1.7\times 10^4)\times(8.5\times 10^{-2})=(1.7\times 8.5)\times(10^4\times 10^{-2})


Recall this product law of indices


a^m\times a^n=a^{m+n}


we apply this law to obtain,


(1.7\times 10^4)\times(8.5\times 10^{-2})=14.45\times10^{4+-2}



This simplifies to


(1.7\times 10^4)\times(8.5\times 10^{-2})=14.45\times10^{2}


We need to rewrite this in standard form;


(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^1\times10^{2}


We apply the product law again to get


(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^{1+2}


This simplifies to

(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^{3}



ANSWER TO QUESTION 2


(6.8\times 10^2)\times(1.3\times 10^{-3})


We rewrite to obtain;


(6.8\times 10^2)\times(1.3\times 10^-3)=(6.8\times 1.3)\times(10^2\times 10^{-3})


Recall this product law of indices


a^m\times a^n=a^{m+n}


we apply this law to obtain,


(6.8\times 10^2)\times(1.3\times 10^-3)=8.84\times10^{2+-3}



This simplifies to


(6.8\times 10^2)\times(1.3\times 10^-3)=8.84\times10^{-1}


This is already in standard form.







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​Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.
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The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

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We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

x^{3}+7 x^{2}+12 x=0

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Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

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we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

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By substituting the values of a,b and c in the quadratic equation we get;

\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}

<em><u>Therefore, the two roots are:</u></em>

\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}

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\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

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