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3 years ago

How do you solve this math problem? 5 = 12 + 4

2 answers:

7 0

You add 12 to 4 to get 16 and then you discover that this is not an equation because the left and right side of the equals sign are not the same

Dennis_Churaev [7]3 years ago

5 0

It’s called a “false equation” bc their is no way to solve it. 16 does NOT equal 5.

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Simplify the rational expression by rationalizing the denominator √15/√6x

noname [10]

Answer:

√10/2x

Step-by-step explanation:

Multiply by √6 on the top and bottom to rationalize the denominator.

(√15×√6)/(√6x×√6) = √90/6x = (√9×√10)/6x = 3√10/6x = √10/2x

4 0

3 years ago

I need help on 23 !!

Marrrta [24]

Answer:

Step-by-step explanation:

1/4 * 2/3 = 2 / 12 = 1/6

3 0

2 years ago

Help plz 6yd - 2ft 7in

Arlecino [84]

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6 0

3 years ago

The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.

tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that perpendicular lines have negative reciprocal slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

$\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}$

so, it passes through the midpoint of AB,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)$

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$

4 0

3 years ago

16.9÷(-1.3) <img src="https://tex.z-dn.net/?f=16.9%20%5Cdiv%20%28%20-%201.3%29" id="TexFormula1" title="16.9 \div ( - 1.3)" a

sattari [20]

16.9 ÷ (-1.3)

= -13

God bless!

7 0

3 years ago