the product of two consecutive odd intergers is one less than six times their sum. find the intergers
The length of a picture frame is one less inch than three times its width. The area of the picture frame is 70 square inches find the dimensions of the picture frame.
20a^2-36a
The hypotenuse of a right tringle is 2 inches longer than the longer leg. The longer leg is 4 inches more that twice the shorter leg. Find the length of the longer leg.
the product of two consecutive odd intergers is one less than six times their sum. find the intergers
n(n+2) = 6(n+n+2)-1. Solve for n.
The length of a picture frame is one less inch than three times its width. The area of the picture frame is 70 square inches find the dimensions of the picture frame.
lw = 70, l = 3w -1 -----> (3w-1)w = 70. Solve for w and use the equation for area to get the length.
20a^2-36a
wat
The hypotenuse of a right tringle is 2 inches longer than the longer leg. The longer leg is 4 inches more that twice the shorter leg. Find the length of the longer leg.
x^2 + (2x+4)^2 = y^2 y = 2x +6
Plus y into the first equation and solve for x to get the length of the short leg, then use the relation the problem gives to get the short leg.
Could someone help me out with this? I just need to find the z-score first for these two things to find the standard deviation but it won't show me how the fuck to do it when I click "show me how to do this".
Here's the problem:
While only 5% of babies have learned to walk by the age of 11 months, 65% are walking by 14 months of age. If the age at which babies develop the ability to walk can be described by a Normal model, find the parameters (mean and standard deviation).
Comments
2^-3 = 1/(2^3) = 1/8
So yeah.....the sign was wrong on the first one, but the second one is definitely 1/8......
-3^-2 = 1/(-3^2) = 1/(-3*-3) = 1/9, for a more comprehensive look.
The length of a picture frame is one less inch than three times its width. The area of the picture frame is 70 square inches find the dimensions of the picture frame.
20a^2-36a
The hypotenuse of a right tringle is 2 inches longer than the longer leg. The longer leg is 4 inches more that twice the shorter leg. Find the length of the longer leg.
y = 2x +6
Plus y into the first equation and solve for x to get the length of the short leg, then use the relation the problem gives to get the short leg.
n^2 + 2n = 6n +6n + 12 - 1 ---> n^2 -10n - 11 = 0 ----> (n-11)(n+1) = 0 ----> n = 11, -1
If you try n= -1, it doesn't work, so n = 11.
lw = 70, l = 3w -1 -----> (3w-1)w = 70. Solve for w and use the equation for area to get the length.
3w^2 - w - 70 = 0. Use the quadratic formula to get w. If you get a negative number, discard it because width can't be negative. Then l = 70/w
Last one.....
x^2 + (2x+4)^2 = y^2
y = 2x +6
x^2 + (2x+4)(2x+4) = (2x+6)(2x+6) -----> x^2 + 4x^2 + 16x + 16 = 4x^2 + 24x + 36-----> x^2 - 8x - 20 = 0 ---> (x-10)(x+2) = 0
x = 10 because you can't have a leg with negative length. Then y = 2(10) + 6 = 26.
x^3-2xy^2=0
x+(2x+4)^2= (2x+6)^2
20a^2 - 36a
12x^2-7xy-12y^2
Here's the problem:
While only 5% of babies have learned to walk by the age of 11 months, 65% are walking by 14 months of age. If the age at which babies develop the ability to walk can be described by a Normal model, find the parameters (mean and standard deviation).
So, I have to find an equation of the tengent line to the graph of the function at the given point. my function is f(x)=e^(1-X) and my point is (1,1)
I'm pretty sure the derivative of that would be -e^(1-X). Assuming I'm correct, where do I go from there?
The formula for the tangent line is y - f(x_0) = f'(x_0)(x-x_0)
So in this case it would be y-f(1) = f'(1)(x-1)
One is the integral of (e^-x)/(1+e^-x)
Other is e^x√(1-e^x)
wut do?