In the expression a=k/b, what is k?

Mathematics

1 answer:

romanna [79]3 years ago

6 0

Answer is k=ba because multiply b on both sides to get rid of it as a denominator.

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The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.

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3 years ago

What is halfway between 8.270 and 8.280

GuDViN [60]

8.275.
If you add both and divide by two you get 8.275

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3 years ago

3x + 4x + 5x = <br><br> May you please help I don’t understand <br> May you show the work too?

Rzqust [24]

The answer is 12x because just add the three numbers and plug in x at the end of the answer

7 0

3 years ago

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Find the probability that a point chosen at random will lie in the shaded area.

son4ous [18]

I think the answer is D

6 0

3 years ago

Given: AB tangent at D, AD = OD = 4 Find: Area of the shaded region

MrMuchimi

Answer:

1.72

Step-by-step explanation:

AB tangent at D, AD = OD = 4

so triangle OAD is right angle with side of 4 and 4.

area of OAD = 1/2 * 4 * 4 = 8

Angle AOD = DAO = 45 deg.

so circular sector OCD area = area of circle O * 45/360

= pi * 4 * 4 * 45/360

= 2pi

Shade area ACD = trigangle OAD - circular sector OCD

= 8 - 2pi

= 1.72

4 0

3 years ago