49–2ax –a^2–x^2
Extract the negative sign
= 49 - (a^2 +2ax +x^2)
Factor using a^2 + 2ab + b^2 = (a+b)^2
= 49 - (a+x)^2
Factor using a^2 - b^2 = (a-b)(a+b)
= ( 7- (a + x)) * (7 + (a + x))
= ( 7- a - x) * (7 + a + x)
The answer is 221 I did the work
Given that the triangle is dilated by factor 3, the image will be found as follows;
The object is at:
A(-7,-3), B(-3,-2), C(-4,-5)
when the image was enlarged the new coordinates will be:
A'=3(-7,-3)=(-21,-9)
B'=3(-3,-2)=(-9,-6)
C'=3(-4,-5)=(-12,-15)
since the image is centered at the point (-7,-6), the final point will be at:
A"=[(-21+-7),(-9+-6)=(-28,-15)
B''=[(-9+-7),(-6+-6)]=(-16,-12)
C''=[(-12+-7),(-15+-6)]=(-19,-17)
thus the coordinates of the final image are:
A''(-28,-15),B''(-16,-12),C''(-19,-17)
Answer:428 7/100
Step-by-step explanation:
Answer: Choice C) 124 square cm
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Explanation:
Let's calculate the area of the trapezoid shown
b1 and b2 are the parallel bases; h is the height of the 2D trapezoid
b1 = 2
b2 = 5
h = 1.5
A = h*(b1+b2)/2
A = 1.5*(2+5)/2
A = 1.5*7/2
A = 10.5/2
A = 5.25
The area of one 2D trapezoid is 5.25 sq cm
There are two of these trapezoids that form the base faces of the trapezoidal prism. So the total base area is 2*5.25 = 10.5 sq cm
Keep this value (10.5) in mind. We'll use it later.
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Now onto the lateral surface area (LSA)
It turns out that the formula for the LSA is
LSA = p*d
where
p = perimeter of the trapezoid shown
d = depth or height of the 3D trapezoid (I'm not using h as it was used earlier)
This formula works for any polygonal base. It doesn't have to be a trapezoid.
In this case the perimeter is,
p = 1.7+2+2.65+5
p = 11.35
So
LSA = p*d
LSA = 11.35*10
LSA = 113.5
Add this LSA to the base area found earlier
10.5+113.5 = 124
The total surface area is 124 square cm
