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lbvjy [14]
2 years ago
12

Which function models the area of a rectangle with side lengths of 2x – 4 units and x + 1 units? What is the area when x = 3?

Mathematics
1 answer:
Bingel [31]2 years ago
7 0

Answer:

8

Step-by-step explanation:

Length = 2x-4

Width = x+1

Let x=3 and find the length and width

Length = 2*3-4 = 6-4 =2

Width = x+1= 3+1 =4

The  area is given by

A = l*w = 2*4 = 8

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How can I this math equation -5 -x =-8
Liula [17]

Answer: x = 3

Step-by-step explanation:

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6 0
2 years ago
How do I do 8b(ii) ? Please help me thank you!
Mila [183]
Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2      Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2    Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


7 0
3 years ago
Which choices are real numbers. Check all that apply. A. (-1024)^1/4. B. (-131072)^1/16. C. (-256)^1/9. D. (-531441)^1/13
vredina [299]

Answer:

B

Step-by-step explanation:

That is my final answer so I maybe wrong so don't take my word for it

4 0
3 years ago
Read 2 more answers
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