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WARRIOR [948]
3 years ago
12

- 2 = 1 \div 3x - 4" alt="x \div 3x - 2 = 1 \div 3x - 4" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Maurinko [17]3 years ago
4 0

Hello from MrBillDoesMath!

Answer:

x = 1/2 (1 +\- i sqrt(23))

Discussion:

x \3x - 2 =   (x/3)*x - 2   =  (x^2)/3 - 2     (*)

1 \3x - 4   =  (1/3)x - 4                               (**)

(*) = (**)   =>

(x^2)/3 -2 = (1/3)x - 4        => multiply both sides by 3

x^2 - 6 = x - 12                 => subtract x from both sides

x^2 -x -6 = -12                   => add 12 to both sides

x^2-x +6  = 0

Using the quadratic formula gives:

x = 1/2 (1 +\- i sqrt(23))

Thank you,

MrB

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How would you go about finding the area and perimeter of a composite figure?
fgiga [73]

Explanation:

The area is the sum of the areas of the non-overlapping parts. The figure is called "composite" because it is composed of figures whose area formulas you know. Decompose the figure into those, find the area of each, then sum those areas to find the area of the whole.

<u>For example</u>

If the figure consists of a rectangle and semicircle, find the areas of each of those. Then add the areas together to find the total area.

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Likewise, the perimeter of a composite figure will be the sum of the "exposed" perimeters of the parts. (Some edges of the figures making up the composition will be internal, so do not count toward the perimeter of the composite figure.)

<u>For example</u>

If the curved edge of the semicircle of the figure described in the example above is part of the perimeter, then its length will be half the circumference of a circle. If the straight edge of the semicircle is "internal" and not a part of the perimeter, its length (the diameter of the semicircle) may need to be partially or wholly subtracted from the perimeter of the rectangle, depending on the actual arrangement of the composite figure. In other words, add up the lengths of the edges that "show."

_____

<em>Additional comments</em>

In the above, we have described how to add the areas of parts of the figure. In some cases, it can be easier to identify a larger figure, or one that is more "complete", then subtract the areas of the parts that aren't there. For example, an L-shaped figure can be decomposed into two rectangles. Or it can be decomposed into a larger rectangle covering the entire outside dimensions, from which a smaller rectangle is subtracted to leave the L-shape. Depending on how dimensions are shown, one computation or the other may be easier.

Likewise, for the purposes of computing the perimeter, lines of the figure may be rearranged in any convenient way, as long as their total length doesn't change. The L-shape just described will have a perimeter exactly equal to the perimeter of the rectangle that encloses its outside dimensions, for example. You can see this if you move the two lines forming the concave edges.

Familiarity with area formulas can help with area. For example, you know that the area of a triangle is the same as that of a rectangle half the height. Likewise, the area of a trapezoid is the area of a rectangle with the same height and a width equal to the midline of the trapezoid.

5 0
3 years ago
How would I go about solving this problem???
schepotkina [342]
So.. if you take a peek at the picture below

the trunk is really just a half-cylinder on top of a square, with a depth of 2 meters

what's the volume?   well, easy enough, take the volume of the cylinder, then half it
take the volume of the rectangular prism, and then add them both

\bf \textit{volume of a cylinder}\\\\&#10;&#10;\begin{array}{llll}&#10;C=\pi r^2 h\\\\&#10;\textit{half that}\\\\&#10;\cfrac{\pi r^2 h}{2}&#10;\end{array}\qquad &#10;\begin{cases}&#10;r=radius\\&#10;h=height\\&#10;-------\\&#10;r=\frac{1}{2}\\&#10;h=2&#10;\end{cases}\implies \cfrac{C}{2}=\cfrac{\pi \left(  \frac{1}{2}\right)^2 2}{2}\\\\&#10;-----------------------------\\\\&#10;\textit{volume of a square}\\\\&#10;V=lwh\qquad &#10;\begin{cases}&#10;l=length\\&#10;w=width\\&#10;h=height&#10;----------\\&#10;l=1\\&#10;w=1\\&#10;h=2&#10;\end{cases}\implies V=2



now.. for the surface area... \bf \textit{surface area of a cylinder}\\\\&#10;\begin{array}{llll}&#10;S=2\pi r(h+r)\\\\&#10;\textit{half that}\\\\&#10;\cfrac{2\pi r(h+r)}{2}&#10;\end{array}\begin{cases}&#10;r=radius\\&#10;h=height\\&#10;-------\\&#10;r=\frac{1}{2}\\&#10;h=2&#10;\end{cases}

now.. for the surface area of the prism... well

is really just 6 rectangles stacked up to each other at the edges

so... get the area of the lateral rectangles, and the one at the bottom, skip the rectangle atop, because is the one overlapping the cylinder, and is not outside, and thus is not surface area then

for the lateral ones, you have a front of 1x1, a back of 1x1 and a left of 1x2 and a right of 1x2, and then the one at the bottom, which is a 1x2

then add both surface areas, and that's the surface area of the trunk

5 0
3 years ago
3x+2y=3y-2 line 1 x+y=10 line 2 solve for x/y
GrogVix [38]

Answer:

x/y=1/4

Step-by-step explanation:

3x+2y=3y-2

3x=3y-2y-2

3x=y-2

y=3x+2

x+y=10

x+3x+2=10

4x+2=10

4x=10-2

4x=8

x=8/4

x=2

2+y=10

y=10-2

y=8

x/y=2/8=1/4

3 0
3 years ago
What is the value of y?
Slav-nsk [51]
In a triangle all of the angles always add up to 180, so let's set up an equation.

79 + 37 + y = 180
Simplify
116 + y = 180
Subtract 116 from both sides.
y = 64

The value of y is 64.

Hopefully this helps! If you have any more questions or don't understand, please comment or DM me, and I'll get back to you ASAP. :)
6 0
3 years ago
The shoe store expected to sell at least 95 pairs of shoes one weekend, but they only sold 77 pairs. What was the approximate pe
Gwar [14]
The error of percentage is the amount of error over the original amount.

In this case it is 22/95  (not 22 over 77 because 77 is not the original amount they expected to sell.   

22/95 = .2315  or 23.2% to the nearest tenth of a percent
Perhaps your teacher just wants 23% as an the approximate error.
7 0
3 years ago
Read 2 more answers
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