Salut. BC²=AC²-AB² , BC=√((10)²-(8)²) , BC=√(100-64) , BC=√36 , BC=6cm. J'espere t,avoir aidé
Answer:
subtract 11 from both sides I think is what you're asking
Step-by-step explanation:
Answer:
4x-5y=3
-3=4x-5y
Step-by-step explanation:
4x-5y = 3
4x-5y = -3
both statements have different answers but has the same problem making the statement x=x and 0=0 making it have no solution
First we have to put the numbers in order
2,6,8,13,16,28,32 and mark the two middle numbers
divide each half into half again and mark the middle number, in this case
2,6,8,13,16,28,32
subtract the 28 from the 6 and get 22
thats the interquartile range
Although the question is not given,I can infere that you need to know the value of x, that is the widht of the frame.
Note that the dimensions of the rectangle covered by the frame are:
8 + 2x and 7 + 2x, meaning that the area covered by the whole figure is:
(8 + 2x) * (7 + 2x).
While the area of the picture is 8 in * 7 in = 56 in^2.
Then, you can state that the area of just the frame meets this equation:
area of the frame = area of the whole figure - area of the picture
34 = (8+2x)(7+2x) - 56
Now solve the equation, starting by expanding the product of the two parenthesis:
8*7 + 16x + 14x + 4x^2 - 56 = 34
4x^2 + 30x + 56 - 56 - 34 = 0
4x^2 + 30x - 34 = 0
Divide by 2 => 2x^2 + 15x - 17 = 0
Factor: (2x - 2)(x + 17/2) = 0
=> 2x - 2 = 0 => x = 1, and x + 17/2 = 0 => x = - 17/2.
Only x = 1 has a real physical meaning so, the solution is x = 1, which is the width of the frame.
Answer: the widht of the frame is 1 inch.
You can check: (8 + 2) (7 + 2) = (10)(9) = 90 = area of picture + the frame.
90 - 56 = 34 which is the area of the frame.
