Answer:the perimeter of the rectangle =A) (−14w+14v) centimeters
Step-by-step explanation:
Step 1
The perimeter of a Rectangle is given as 2 (l+ w)
If Length = 2v+3w centimeters.
and width =5v-10wcentimeters
Then Perimeter =2 (2v+3w + 5v-10w)
Perimeter=2(7v -7w)
Perimeter= 14v - 14w or can be written as -14w +14v
Step-by-step explanation:
part A:
ABCD is transformed to obtain figure A′B′C′D′:
1) by reflection over x-axis, obtain the image :
A(-4,-4) B(-2,-2) C(-2, 1) D(-4, -1)
2) by translation T (7 0), obtain the image :
A'(3,-4) B'(5,-2) C'(5, 1) D'(3, -1)
part B:
the two figures are congruent.
the figures that transformed by reflection either or translation will obtain the images with the same shape and size (congruent)
Answer:
- x = log(y/4)/log(1.0256)
- your answer for y=12 is correct
Step-by-step explanation:
The question is asking you to solve ...
y = f(x)
for x. (In other words, find the inverse function.)
You already did this using a constant for y. Do the same thing with y instead of the constant.
y = 4(1.0256^x)
y/4 = 1.0256^x . . . . . . . divide by 4
log(y/4) = x·log(1.0256) . . . . . take logs
log(y/4)/log(1.0256) = x . . . . . divide by the coefficient of x
Now, you have a model for x in terms of y, which is what the question is asking for.
x = log(y/4)/log(1.0256) . . . . . . . exact expression
When y=12, this is ...
x = log(12/4)/log(1.0256) ≈ 43.46 . . . . weeks
_____
This is a linear equation in log(y), so can be written as such:
x = 91.0912·log(y) -54.8424 . . . . . approximate expression
The two cities are actually 427.5 miles apart.
Part (a)
<h3>Answer: y1 and y3 are perpendicular</h3>
This is because the two slopes 2 and -1/2 multiply to -1. Perpendicular slopes multiply to -1 assuming neither line is vertical or horizontal.
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Part (b)
Graph each line to see where they cross. The three points of intersection are
(0,4)
(2,-2)
(4,2)
The order of the points doesn't matter.
You could also form three systems of equations pairing up the equations, and solving each system. That way you can find the points of intersection. Graphing may be a better and faster route in my opinion. See the diagram below.