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BartSMP [9]
3 years ago
6

25 POINTS + BRAINLIEST IF YOU EXPLAIN ANSWER

Mathematics
2 answers:
ella [17]3 years ago
5 0

Answer:

4th degree

Step-by-step explanation:

We can use a technique called finite differences.

In column 1  put the differences  between the y values  (y2-y1)

Keep adding columns with the differences until they become constant.

Column 2 is column1 differences  (-2- -30) etc

Column3 is column 2 differences

Column 4 is column3 differences

The differences in column4 are the same, so we can stop

This will be a 4th degree polynomial

x    y          column 1               column 2      column 3        column 4

-2  30        

-1   0                 -30                    

0   -2               -2                           28

1      0               2                           4               -24

2     30           30                          28               24              48

3    160           130                        100               72               48

4      510        350                       220               120             48

Rina8888 [55]3 years ago
4 0

Answer:

4th degree ...........

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7 0
3 years ago
If the equation y² - (K-<br>2y + 2k +1 = 0<br>with equal roots find the value of k​
777dan777 [17]

Answer:

y² - (K-  2)y + 2k +1 = 0

equal roots means D=0

D= b^2 - 4ac

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so,

(k-2)^2 - 4(1)(2k+1) = 0

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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
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