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Maru [420]
3 years ago
9

Are these two triangles congruent? *

Mathematics
2 answers:
Ahat [919]3 years ago
8 0

Answer: yes

hope it helped if it did don't forget to drop a heart

Step-by-step explanation:

I am Lyosha [343]3 years ago
7 0

Answer:

yes  SAS

Step-by-step explanation:

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Two 2 1/2 inch plastic strips and two 5 1/3 inch plastic strips are used to form a rectangle. What is the perimeter of the recta
svp [43]
In this question the given information's should be closely noted. The length and width of the perimeter are already given. Based on those information's the answer to the question can be easily deduced.
Length of the rectangle = 2 1/2 inch
                                       = 5/2 inch
Width of the rectangle = 5 1/3 inch
                                         = 16/3 inch
Then
Perimeter of a rectangle = 2 ( Length + Width)
                                        = 2 [(5/2) + (16/3)]
                                         = 2 [ (45 + 32)/6]
                                         = 2 * (77/6)
                                         = 77/3 inch
                                         = 25 2/3 inch
So the perimeter of the rectangle in question is 25 2/3 inch. I hope the procedure is clear to you.
5 0
3 years ago
Read 2 more answers
Use the quadratic formula to find the solutions for..<br> y = 2x^2 - 9x + 5.
Yakvenalex [24]

Answer: x=\frac{9}{4}-\frac{\sqrt{41} }{4}  \\x=\frac{9}{4}+\frac{\sqrt{41} }{4}

Step-by-step explanation:

2x^2-9x+5

a=2

b=-9

c=5

x=\frac{-b\frac{+}{}\sqrt{b^2-4ac}  }{2a}

x=\frac{-(-9)\frac{+}{}\sqrt{(-9)^2-4(2)(5)}  }{2(2)}

x=\frac{9\frac{+}{}\sqrt{81-40}  }{4}

x=\frac{9\frac{+}{}\sqrt{41}  }{4}

x=\frac{9}{4}-\frac{\sqrt{41} }{4}  \\x=\frac{9}{4}+\frac{\sqrt{41} }{4}

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3 years ago
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Marta_Voda [28]
Step 1 , switch sides - 29-3b<10
step 2 , subtract 29 from both sides - -3b< -19
step 3 , multiply both sides by -1 - 3b > 19
step 4 (answer) - divide by 3 = 19/3
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3 years ago
In a board game, students draw a number do not replace it and then draw a second number
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3 years ago
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