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3 years ago

In the diagram, the areas of ΔADC and ΔDCB are in a ratio of 3 : 4. What are the coordinates of point C?

Mathematics

2 answers:

Kisachek [45]3 years ago

5 0

(1/2(AC) * DC) / (1/2(BC) * DC) = 3/4
AC / BC = 3/4

The coordinate of C be (x, y), then
((4(1) + 3(8))/7, (4(-9) + 3(-2))/7) = ((4 + 24)/7, (-36 - 6)/7) = (28/7, -42/7) = (4, -6)

anygoal [31]3 years ago

5 0

I hope this helps you

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$y = \frac{5000r}{3} -50$

Step-by-step explanation:

Represent the SAT score with y and the rate with r.

So, we have:

$(r_1,y_1) = (90\%,1450)$

$(r_2,y_2) = (69.6\%,1110)$

Required

Determine the equation in slope intercept form

First, we calculate the slope

$m =\frac{y_2 - y_1}{r_2 - r_1}$

This gives:

$m =\frac{1110 - 1450}{69.6\% - 90\%}$

$m =\frac{-340}{-20.4\%}$

Convert percentage to decimal

$m =\frac{-340}{-0.204}$

$m =\frac{340}{0.204}$

Multiply by 1000/1000

$m =\frac{340*1000}{0.204*1000}$

$m =\frac{340000}{204}$

$m = \frac{5000}{3}$

The equation is then calculated as:

$y - y_1 = m(r - r_1)$

This gives:

$y - 1450 = \frac{5000}{3}(r - 90\%)$

Open Bracket

$y - 1450 = \frac{5000r}{3} - \frac{5000}{3}*90\%$

Convert percentage to decimal

$y - 1450 = \frac{5000r}{3} - \frac{5000}{3}*0.90$

$y - 1450 = \frac{5000r}{3} - 5000*0.30$

$y - 1450 = \frac{5000r}{3} - 1500$

Make y the subject

$y = \frac{5000r}{3} - 1500+1450$

$y = \frac{5000r}{3} -50$

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2 years ago