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hodyreva [135]
3 years ago
12

Could someone help me with this? it'd be very appreciated!

Mathematics
1 answer:
Serga [27]3 years ago
8 0

Answer:

47,759

Step-by-step explanation:

77 + 216 x 326 then divide by 2

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The steps to derive the quadratic formula are shown below:
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Answer:

  (c)  x plus b over 2 times a equals plus or minus the square root of the quantity b squared minus 4 times a times c all over the square root of 4 times a squared

Step-by-step explanation:

The next step is to take the square root of both sides of the equation. It can help to show the intermediate steps.

<h3>Result so far</h3>

The last step shown in the derivation so far is ...

  x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=-\dfrac{4ac}{4a^2}+\dfrac{b^2}{4a^2}

<h3>Next step</h3>

The left side of the above expression can be written as a square, and the right side can be written over one denominator. Then the square root is taken as the next step.

  \left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}\\\\\sqrt{\left(x+\dfrac{b}{2a}\right)^2}=\sqrt{\dfrac{b^2-4ac}{4a^2}}\\\\\boxed{x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}}\qquad\text{"next step"}

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pickupchik [31]

Step-by-step explanation:

<em>Hello</em><em>,</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

<em>Let</em><em> </em><em>ABC</em><em> </em><em>be</em><em> </em><em>a Right angled triangle</em><em>, </em>

<em>where</em><em>,</em><em> </em><em>AB</em><em> </em><em>=</em><em> </em><em>3</em>

<em>BC</em><em>=</em><em> </em><em>1</em><em>0</em>

<em>and</em><em> </em><em>AC</em><em>=</em><em> </em><em>x</em>

<em>now</em><em>,</em>

<em>As</em><em> </em><em>the</em><em> </em><em>triangle</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>Right angled triangle</em><em>, </em><em>taking</em><em> </em><em>angle C</em><em> </em><em>as</em><em>refrence</em><em> </em><em>angle</em><em>.</em><em> </em><em>we</em><em> </em><em>get</em><em>,</em>

<em> </em><em>h</em><em>=</em><em> </em><em>AC</em><em> </em><em>=</em><em> </em><em>x</em>

<em>p</em><em>=</em><em> </em><em>AB</em><em> </em><em>=</em><em> </em><em>3</em>

<em>b</em><em>=</em><em> </em><em>BC</em><em>=</em><em> </em><em>1</em><em>0</em>

<em>now</em><em>,</em><em> </em><em>by</em><em> </em><em>Pythagoras</em><em> </em><em>relation we get</em><em>, </em>

<em>h =    \sqrt{ {p}^{2} +  {b}^{2}  }</em>

<em>or ,\: h =  \sqrt{ {3}^{2}  +  {10}^{2} }</em>

<em>by</em><em> </em><em>simplifying it we get</em><em>, </em>

<em>h</em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em>.</em><em>4</em><em>4</em><em>0</em><em>3</em><em>0</em>

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Dan drew the line of best fit on the scatter plot shown below: A graph is shown with scale along x axis from 0 to 10 at incremen
Anit [1.1K]
     0, 3
-   10, 15
= -10, -12
therefore, the slope is 6/5, and the intercept (c) is as supplied, 3.
the equation, y=mx+c or y = a + bx, can be applied here where m or b = 6/5, and a or c = 3.

therefore the equation is y=6/5x+3.

To test this, you can put in y = 10(6/5)+3, which spits out y = 15. This way we know it *should* work.

3 0
3 years ago
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