Answer: Any two polygons are similar if their corresponding angles are congruent and the measures of their corresponding sides are proportional: In the figure above the ratio or the scale factor of the quadrilateral to the left versus the quadrilateral to the right is ½.

Step-by-step explanation: This what I could coem up with. Hope this helps !!

5 0

3 years ago

Please help me with this problem i tried

mezya [45]

Answer:

I think the answer is 1280

4 0

3 years ago

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the length of a rectangle is 6in.more than than its width .Its area is 50 sq.in Find the width of the rectangle....

yulyashka [42]

L=6+W
A=LW
sub 6+W for L
A=(6+W)(W)
expand
oh, A=50
50=W²+6W
minus 50 both sides
0=W²+6W-50
we gots to use quadratic formula
for
aW²+bW+c=0
W= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
we got
a=1
b=6
c=-50

W= $\frac{-(6)+/- \sqrt{(6)^2-4(1)(-50)} }{2(1)}$
W= $\frac{-6+/- \sqrt{36+200} }{2}$
W= $\frac{-6+/- \sqrt{236} }{2}$
W= $\frac{-6+/- 2\sqrt{59} }{2}$
W= $-3+/- \sqrt{59}$

W= $-3+ \sqrt{59}$ or W= $-3- \sqrt{59}$
aprox
W=-10.6811 or W=4.68115
can't be negative width

width=4.68115 or -3√59

7 0

3 years ago

Cindy is preparing for a 25-mile charity run hosted by her school's Athletic Club on their home track. In preparation for the ru

deff fn [24]

What are the answer choices?

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3 years ago

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