In which quadrant(s) do the graphs of y = x^2 and y = 3x +10 intersect?

Mathematics

1 answer:

vichka [17]2 years ago

5 0

Answer:

Step-by-step explanation:

y=3x+10

y=x²

x²-3x-10=0

x²-5x+2x-10=0

x(x-5)+2(x-5)=0

(x-5)(x-2)=0

x=5,2

when x=5,y=5²=25

point of intersection is (5,25)

when x=2

y=2²=4

so point of intersection is (2,4)

which shows that they intersect in first quadrant.

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WILL GIVE A BRAINLIEST

skelet666 [1.2K]

Find slope
slope betwen 2 points (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)

so
first make denoms same
denom is 10
(4/5,1/5) turns to (8/10,2/10)
(1/2,3/2) turns to (5/10,15/10)
slope=(15/10-2/10)/(5/10-8/10)=(13/10)/(-3/10)=13/-3=-13/3

use point slope
use (4/5,1/5)
y-y1=m(x-x1)
a point is (x1,y1) and slope is m
y-1/5=-13/3(x-4/5)
expand
wait, we don't need to do that, the only one with slope -13/3 is D
answer is D

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Find the value of x and y

Romashka [77]

Step-by-step explanation:

$\triangle ABC \sim\triangle DEF..(Given) \\ \therefore \frac{AB}{DE} =\frac{BC}{EF}=\frac{AC}{DF}..(CSST)\\\\\therefore \frac{6}{4} =\frac{7}{x}=\frac{8}{y}\\\\ \therefore \frac{6}{4} =\frac{7}{x} \\ \therefore \: x = \frac{7 \times 4}{6} \\ \\\therefore \: x = \frac{28}{6} = \frac{14}{3} \\ \\ \huge \red{ \boxed{ \therefore \: x = \frac{14}{3} }} \\ \\ \because \: \frac{6}{4} =\frac{8}{y} \\ \\ \therefore \:y = \frac{4 \times 8}{6} \\ \\ \therefore \:y = \frac{32}{6} \\ \\ \huge \purple{ \boxed{ \therefore \:y = \frac{16}{3} }}\\ \\$