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kari74 [83]
3 years ago
15

carpenter has a board that is 8 feet long. e cuts off two pieces. One piece is 32 feet ug and the other is 2 feet long. How ch o

f the board is left?​
Mathematics
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

2 1/6 feet

Step-by-step explanation:

A carpenter has a board that is 8 feet long. He cuts off two pieces. One piece is 3 1/2 feet long and the other is 2 1/3 feet long. How much of the board is left?

Total length of the board = 8 feet

Piece A = 3 1/2 feet

Piece B = 2 1/3 feet

Piece A + piece B

= 3 1/2 feet + 2 1/3 feet

= 7/2 + 7/3

= (21+14) / 6

= 35/6

= 5 5/6 feet

Length of the board remaining = Total length of the board - length of the two pieces

= 8 feet - 35/6 feet

= (48-35) / 6

= 13/6 feet

= 2 1/6 feet

Length of the board remaining = 2 1/6 feet

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Write three consecutive multiples of 13 in a general form.
S_A_V [24]

Answer:

answer given in step by step...

Step-by-step explanation:

13x+13(x+1)+13(x+2)=312

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x=7

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30 POINTS PLEASE HELP LOOK AT THE PICTURES ITS EASY IM JUST DUMB!!!! Which stem-and-leaf plot represents the data sets shown?
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Step-by-step explanation:

6 0
3 years ago
A student was given the line y=2/3x-1 and the point (-7, 1/2) and asked to find an equation that went through the given point an
Colt1911 [192]

keeping in mind that perpendicular lines have <u>negative reciprocal</u> slopes,  hmmm what's the slope of y=2/3x-1 anyway?


\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{7em}y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x-1 \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}


so, notice, "one of his mistakes" is that he used 3/2 as the slope, not -3/2.

so, we're really looking for a line whose slope is -3/2 and runs through (-7, 1/2).


\bf (\stackrel{x_1}{-7}~,~\stackrel{y_1}{\frac{1}{2}})~\hspace{10em} slope = m\implies -\cfrac{3}{2} \\\\\\ \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-\cfrac{1}{2}=-\cfrac{3}{2}[x-(-7)]\implies y-\cfrac{1}{2}=-\cfrac{3}{2}(x+7) \\\\\\ y-\cfrac{1}{2}=-\cfrac{3}{2}x-\cfrac{21}{2}\implies y=-\cfrac{3}{2}x-\cfrac{21}{2}+\cfrac{1}{2}\implies y=-\cfrac{3}{2}x-10

3 0
3 years ago
Given H(6, 7) and I(-7,-6), if point G lies 1/2 of the way along Hl. John says that point G is located at the origin. Is John co
atroni [7]

Given:

Two points are (6, 7) and I(-7,-6).

Point G lies \dfrac{1}{2} of the way along Hl.

John says that point G is located at the origin.

To find:

Whether John is correct or not.

Solution:

Point G lies \dfrac{1}{2} of the way along Hl. I means point G is the midpoint of HI.

Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

G=\left(\dfrac{6-7}{2},\dfrac{7-6}{2}\right)

G=\left(\dfrac{-1}{2},\dfrac{1}{2}\right)

The location of point G is at \left(\dfrac{-1}{2},\dfrac{1}{2}\right), which is not equal to origin, i.e., (0,0).

Therefore, John is incorrect.

4 0
3 years ago
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