Multiply denominator by whole number then add that number to the nominator. Which would be 5 times 1 plus 3. SO 5+3 IS 8 and you keep the denominator so the first one is 8/5
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 7 − x 2 y=7-x2. what are the dimensions of such a rectangle with the greatest possible area?
9514 1404 393
Answer:
M₅ = 1.6570
Step-by-step explanation:
The basic idea is to find the function values at x ∈ {1.1, 1.3, 1.5, 1.7, 1.9}, add those values, and multiply the result by 1/5. Those x-values are the midpoints of intervals 0.2 units wide, between x = 1 and x = 2. The width 0.2 is the total interval width (2-1=1) divided by n = 5.
The calculation and results are shown in the attachment. Rounded to 4 decimal places, the integral is ...
M₅ = 1.6570
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The actual integral is about 1.6574, so this is pretty close.
The solution of a system of equations is the point at which the two lines intersect. Since y is already isolated, we can set the equations equal to each other by doing:
4x=4x+1
Subtract 4x from both sides
0=1
Since 0 cannot equal 1 in any case, the lines never intersect. There is no solution to the problem.
Answer:
c)-5(3x+2y)(3x-2y)
Step-by-step explanation:
1) lets find out if -45 and 20 have a GCF
-45 and 20 are both dividable by -5 so
-5(9x^2-4y^2)
1) you can see that 9 and -4 are perfect square so:
-5(3x+2y)(3x-2y)
Recheck:
1) lets use FOIL to see if this is right(if its right we will get back our original equation)
-5(9x^2-6xy+6xy-4y^2)
-5(9x^2-4y^2)
-45x^2+20y^2
hope this helps!