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GaryK [48]
2 years ago
7

Find the area of the polygon

Mathematics
1 answer:
AlekseyPX2 years ago
7 0

Answer:

The answer is 383m^{2} .

To find the area of a rectangle, we multiply the length of the rectangle by the width of the rectangle.

Please Mark Brainliest If This Helped!

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Edward can run 1/2 mile in 300 seconds what is edwards unit rate
Vilka [71]
Rate/ speed = distance / time;
= 1/2mile / 300 seconds;
= 0.001666666..... mile/second
7 0
3 years ago
QUICKLY HELP ME PLSSSSS
joja [24]

Answer:

479

Step-by-step explanation:

342=2 portions

171=1 portion

2+5=7 portions in total

171 times 7=479

hope this helps

please can i get brainliest

7 0
2 years ago
Read 2 more answers
Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (
kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2}) is given as continuous function, so there exist lim_{(x,y)\rightarrow(0,0)}f(x,y) and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})

we know that cos^2A+sin^2A=1, so we have that

f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)

lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19

So f(0,0)=ln19.

8 0
3 years ago
Of the 125 guests invited to a wedding, 100 attended the wedding. What percent of the invited guests attended the wedding?
RSB [31]
100/125 * 100 = 80 % (80 percent)
6 0
3 years ago
A red gumball is 2 5/8 Inches across. A green gumball is 2 5/6 inches across and a blue gumball in 2 7/9 inches across. list the
Masteriza [31]

Answer: 5 8 7 9 5 6

Step-by-step explanation:

Because you go by which is closer to the denominator

8 0
3 years ago
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