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harkovskaia [24]
3 years ago
10

a single box of thumb tacks weighed 3 1/2 if a teacher had 4 1/7 boxes of thumbtacks how much would their combined weight be?

Mathematics
1 answer:
lana66690 [7]3 years ago
6 0
Their combined weight would be 7 2/14. here are the steps of how I got my answer.
1. I made the answers into improper fractions. 3 1/2= 7/2 and 4 1/7= 29/7.
2. then I found the least common multiple (LCM) of 2 and 7 which is 14.
3. next in order to make both denominators become 14 I cant just make 7/2 into 7/14 and 29/7 into 29/14.  so I multiplied 29/7 by 2 which got me 58/14 and 7/2 by 7 which got me 42/14.
4. then I did 42/14+ 58/14= 100/14.
5. Then the last step was to make the improper fraction 100/14 into a proper fraction. Now how do you do that? BY DIVIDING! I divided 100 by 14 and I got my answer of 7 2/14.

Hope this helped! :)
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Answer:

domain - (-infinity,infinity)

range - (-infinity, 2]

axis of sym - x=-1

y int - (0,-1)

at what value (row 1) - (-1,2)

Step-by-step explanation:

domain is side to side, it never ends bc of the arrows

range is top to bottom, there is no bottom bc of the arrow, but it has a top at 2 (the value on the y axis)

the axis of sym is basically the line that runs through the vertex, it's where you could fold it in half and it would still match up

8 0
3 years ago
Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).
lions [1.4K]

Check the picture below.  so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}

(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill

\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}

6 0
2 years ago
Hey i need help can sb help me please
Crazy boy [7]

Answer:

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Step-by-step explanation:

A number minus a negative number will be the same thing as addition

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