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Korvikt [17]
3 years ago
5

I need help someone help me please <3

Mathematics
1 answer:
ElenaW [278]3 years ago
4 0

Answer:

y=-\frac{3}{2}x-2

Step-by-step explanation:

using the y=mx+b equation, where m=slope and b=y-intercept, we can place our known slope and y-intercept to form our answer, y=-\frac{3}{2}x-2 :-) hope this helped!

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haley has $20 in a savings account that earns 10% annually. the interest is not compounded. how much will she have in 2 years
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Answer:

240

Step-by-step explanation:

1 percent=10 dollars

10 percent=100 dollars

=120 dollars in one year

times 2 [2x]=

240

Check:

10 times 10=100 dollars

100+20=120

x2=

120+120 <em>or</em> 120 times 2=

240

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Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from
Lapatulllka [165]
<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units</h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and      y_3=8

The area of a triangle is= \frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]

=|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|

=|-54| = 54 square units

The length of AC = \sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}

                          = \sqrt{(2-12)^{2} +(1-8)^2}

                         =\sqrt{149} units

Let the perpendicular distance from B to AC be = x

According To Problem

\frac{1}{2} \times  x \times \sqrt{149} = 54

⇔x =\frac{108}{\sqrt{149} } units

Therefore the perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units

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3 years ago
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