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stellarik [79]
3 years ago
7

Plz help me because i can't find the number 45 in the graph chat

Mathematics
2 answers:
Delicious77 [7]3 years ago
8 0

Answer:

The answer is H

I can tell because there is a line half between 30 and 60 and 45 is halfway between 30 and 60 so.

Step-by-step explanation:

Karolina [17]3 years ago
3 0
The answer is H because per every 3 parents there is 45 students, 45/3 is 15 and if you look at H y1= x15 and y2=x30 and so n so
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Solve the system of equations using substitution or elimination. Show all work for credit.
Tomtit [17]
Using substitution:
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y =  \frac{ -  \frac{1}{2} x + 2}{3}  \\ y =  - x + 9

Since both expressions are equal to y, you have to equal both expressions like this:
\frac{  - \frac{ 1}{2}x + 2 }{3}  =  - x + 9
Now you can solve the equation:
-  \frac{1}{2} x + 2 = 3( - x + 9) \\   - \frac{ 1}{2}x  + 2 =  - 3x + 27 \\  \\  -  \frac{1}{2} x + 3x = 27 - 2 \\   \frac{5}{2} x = 25 \\ x =  \frac{(25)(2)}{5}  \\ x = 10
Knowing x=10, you can use any of the expressions we found before to find y. In this case I'm going to use y= -x+9 because it's simpler but boy should give you the same result

y =  - (10) + 9 \\ y =  - 1
So, the answer is x=10 and y=-1
4 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}

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Answer:

Ur question is ununderstable

Step-by-step explanation:

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bonufazy [111]
3 and 60 are the extremes while 15 and 12 are the means. 
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