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Sergeu [11.5K]
3 years ago
12

How to solve question question 9. e) ? Including steps are appreciated!

Mathematics
1 answer:
Vanyuwa [196]3 years ago
4 0
What is the question
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Annie is creating a stencil for her artwork using a coordinate plane. The beginning of the left edge of the stencil falls at (−1
ludmilkaskok [199]

Answer: 8 -1

Step-by-step explanation:

referring to brainly.com/question/16983018

The given parameters are;

Location of the beginning of the left edge of the stencil = (-1, 2)

Location of the detail = (2, 1)

Ratio of detail distance from beginning to detail to distance from beginning to stencil end = 1:2

Distance from beginning to detail = √((-1 - 2)² +  (2 - 1)²) = √10

Given that the ratio of the length of the detail to the length of the end after the detail is 1:2 therefore;

√10:Length of stencil side = 1:2

Distance from detail to stencil end = 2×√10

Which gives;

Slope of line = tan⁺¹((2 - 1)/(-1 - 2)) = tan⁺¹(-1/3) = -18.435°

x-coordinates of the end of the stencil = 2√10 × cos(-18.435°) + 2 = 8  

y coordinates of the end of the stencil = 2√10 × sin(-18.435°) + 1 = -1  

The coordinates of the end of the stencil = (8, -1)

5 0
2 years ago
I marked all 3 parts. I need to know part C (the shaded part). anybody mind helping?​
Mnenie [13.5K]

Answer:

Area of shaded part ABCEF = 66 sq.cm

Step-by-step explanation:

AB = 8cm

CD = 8cm

Let DE = x cm

CE = 3x cm

CD =  CE + DE = 8cm

        x + 3x = 8

4x = 8

x = 8/4 = 2 cm

DE = 2cm

CE = 3 * 2 = 6 cm

Area of triangle ADE = 1/2 * base * height

                                   = 1/2 * DE * AD

                                    = 1/2 * 2 * 11 = 11 sq. cm

Area of triangle AEF = Area of triangle ADE = 11 sq. cm

Area of Rectangle ABCD =  l * b = 8 * 11 = 88 sq.cm

Area of shaded part ABCEF =  Area of Rectangle ABCD - (Area of triangle AEF + Area of triangle ADE)

    = 88 - ( 11 + 11 )  = 88 -22 = 66 sq.cm

6 0
3 years ago
Hi gys how stay home stay safe I am new to brainly I don't have and followers I beg you to follow me
Gnesinka [82]
I got you lol have a good day
7 0
3 years ago
Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=3, u'(0)= -4, v(0)= 2, v'(0)= -7. Find the value
3241004551 [841]
\bf \cfrac{d}{dx}[u(x)\cdot v(x)]\implies \cfrac{du}{dx}\cdot v(x)+u(x)\cdot \cfrac{dv}{dx}\quad 
\begin{cases}
u(0)=3\qquad u'(0)=-4
\\\\
v(0)=2\qquad v'(0)=-7
\end{cases}
\\\\
\left[ \cfrac{du}{dx} \right]_{x=0}\cdot v(0)+u(0)\cdot \left[ \cfrac{dv}{dx} \right]_{x=0}

so.. hmm pretty sure you know what that is
7 0
3 years ago
Read 2 more answers
Points A and B have opposite x-coordinates but the same y-coordinates.
n200080 [17]

Answer:

c

Step-by-step explanation:

start before a and count until you get to B

5 0
3 years ago
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