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3 years ago

What is 23/35 simplified?

1 answer:

7 0

There is no least common enominator for this answer due to the fact that these numbers don't share a common factor (23 is odd number with only two factors: 1, 23; while 35 is odd number with 4 factors: 1, 5, 7, 35) other than 1.

*But if you do 1/1, it will be 1; so this won't work properly.

So the answer is 23/35.

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A Russian blue kitten costs $150 more than Manx kitten 9 blue kittens and 8 Manx kittens costs. Total of 14,100 write down a alg

Dmitriy789 [7]

Answer:

The equations 9x+8y=14100 and x = y+150 are in the simplest form.

Step-by-step explanation:

Given that:

Russian blue kitten costs $150 more than Manx kitten.

Cost of 9 blue kittens and 8 Manx kittens = $14100

Let,

x be the price of one Russian blue kitten.

y be the price of one Manx kitten.

According to given statement;

9x+8y=14100

x = y+150

Hence,

The equations 9x+8y=14100 and x = y+150 are in the simplest form.

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3 years ago

What is the answer to 11≥3+4u

slega [8]

Answer:

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Step-by-step explanation:

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4 years ago

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victus00 [196]

The slope = y value / corresponding x value

use the point (3,15)

slope = 15 / 3 = 5

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4 years ago

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Darya [45]

Answer:

I think you'll have to graph that

Step-by-step explanation:

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3 years ago

Solve the following biquadratic equation: <img src="https://tex.z-dn.net/?f=%28%287%21%29%2Ax%5E4%29%2B%28%285%21%29%2Ax

jekas [21]

$\large\begin{array}{l} \textsf{Solve the equation}\\\\ \mathsf{7!\cdot x^4+5!\cdot x^2-3=0}\\\\ \mathsf{7!\cdot (x^2)^2+5!\cdot x^2-3=0}\\\\\\ \textsf{Substitute}\\\\ \mathsf{x^2=t\quad(t\ge 0)}\\\\\\ \textsf{so the equation becomes}\\\\ \mathsf{7!\cdot t^2+5!\cdot t-3=0}\quad\Rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=7!}\\\mathsf{b=5!}\\\mathsf{c=-3} \end{array} \right. \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(5!)^2-4\cdot 7!\cdot (-3)}\\\\ \mathsf{\Delta=(5!)^2+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7!}\\\\ \mathsf{\Delta=5!\cdot 5!+12\cdot 7\cdot 6\cdot 5!}\\\\ \mathsf{\Delta=5!\cdot (5!+12\cdot 7\cdot 6)}\\\\ \mathsf{\Delta=5!\cdot (120+504)}\\\\ \mathsf{\Delta=5!\cdot 624} \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=120\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3\cdot 5)\cdot 2^4\cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^{3+4}\cdot 3\cdot 5 \cdot 3\cdot 13}\\\\ \mathsf{\Delta=2^7\cdot 3^2\cdot 5 \cdot 13} \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=2^{6+1}\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{6}\cdot 2\cdot 3^2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=2^{3\,\cdot\,2}\cdot 3^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 2\cdot 5 \cdot 13}\\\\ \mathsf{\Delta=(2^3\cdot 3)^2\cdot 130} \end{array}$

$\large\begin{array}{l} \mathsf{t=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-(5!)\pm \sqrt{(2^3\cdot 3)^2\cdot 130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-120\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7!}}\\\\ \mathsf{t=\dfrac{-2^3\cdot 3\cdot 5\pm 2^3\cdot 3\sqrt{130}}{2\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 4\cdot 7\cdot (2\cdot 3)\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{2^3\cdot 3\cdot \big(\!\!-5\pm \sqrt{130}\big)}{2\cdot 2^2\cdot 3\cdot 7\cdot 2\cdot 5\cdot 3!}} \end{array}$

$\large\begin{array}{l} \mathsf{t=\dfrac{(2^3\cdot 3)\cdot \big(\!\!-5\pm \sqrt{130}\big)}{(2^3\cdot 3)\cdot 7\cdot 2\cdot 5\cdot 3!}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{7\cdot 2\cdot 5\cdot 3\cdot 2\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 7\cdot 5\cdot 3\cdot 1}}\\\\ \mathsf{t=\dfrac{-5\pm \sqrt{130}}{2^2\cdot 105}} \end{array}$

$\large\begin{array}{l} \begin{array}{rcl} \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}&~\textsf{ or }~& \end{array} \mathsf{t=\dfrac{-5-\sqrt{130}}{2^2\cdot 105}}\quad\textsf{(useless, since }\mathsf{t\ge 0}\textsf{)}\\\\ \mathsf{t=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}} \end{array}$

$\large\begin{array}{l} \textsf{Substituite back for }\mathsf{x^2=t:}\\\\ \mathsf{x^2=\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}\\\\ \mathsf{x=\pm \sqrt{\dfrac{-5+\sqrt{130}}{2^2\cdot 105}}}\\\\ \mathsf{x=\pm \dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}\\\\ \boxed{\begin{array}{rcl} \mathsf{x=-\,\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}}&~\textsf{ or }~&\mathsf{x=\dfrac{1}{2}\sqrt{\dfrac{-5+\sqrt{130}}{105}}} \end{array}} \end{array}$

$\large\begin{array}{l} \textsf{Solution: }\mathsf{S=\left\{-\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}},\,\frac{1}{2}\sqrt{\frac{-5+\sqrt{130}}{105}}\right\}.} \end{array}$

If you're having problems understanding the answer, try to see it through your browser: brainly.com/question/2094277

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}$

Tags: solve biquadratic equation factorial Bhaskara solution

4 0

3 years ago