we know that
The density is equal to the formula
![Density=\frac{mass}{volume}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7Bmass%7D%7Bvolume%7D)
in this problem we have
![mass=23.401\ gr](https://tex.z-dn.net/?f=mass%3D23.401%5C%20gr)
<u>Find the volume of a piece of green jade</u>
![Volume=56.5\ ml-50.0\ ml=6.5\ ml](https://tex.z-dn.net/?f=Volume%3D56.5%5C%20ml-50.0%5C%20ml%3D6.5%5C%20ml)
Substitute the values in the formula of density
![Density=\frac{23.401}{6.5}](https://tex.z-dn.net/?f=Density%3D%5Cfrac%7B23.401%7D%7B6.5%7D)
![Density=3.60\frac{gr}{ml}](https://tex.z-dn.net/?f=Density%3D3.60%5Cfrac%7Bgr%7D%7Bml%7D)
therefore
<u>the answer is</u>
![Density=3.60\frac{gr}{ml}](https://tex.z-dn.net/?f=Density%3D3.60%5Cfrac%7Bgr%7D%7Bml%7D)
Answer:
D) ![\frac{4x-3} {x+6}](https://tex.z-dn.net/?f=%5Cfrac%7B4x-3%7D%20%7Bx%2B6%7D)
Step-by-step explanation:
First we have to factorize and then cancel the like terms.
Given,
![\frac{4x^2-7x+3} {x^{2}+5x-6}\\=\frac{4x^2-3x-4x+3} {x^{2}+6x-x-6}\\=\frac{x(4x-3)-1(4x-3)} {x(x+6)-1(x+6)}\\=\frac{(4x-3)(x-1)} {(x+6)(x-1)}](https://tex.z-dn.net/?f=%5Cfrac%7B4x%5E2-7x%2B3%7D%20%7Bx%5E%7B2%7D%2B5x-6%7D%5C%5C%3D%5Cfrac%7B4x%5E2-3x-4x%2B3%7D%20%7Bx%5E%7B2%7D%2B6x-x-6%7D%5C%5C%3D%5Cfrac%7Bx%284x-3%29-1%284x-3%29%7D%20%7Bx%28x%2B6%29-1%28x%2B6%29%7D%5C%5C%3D%5Cfrac%7B%284x-3%29%28x-1%29%7D%20%7B%28x%2B6%29%28x-1%29%7D)
Now, we got the like term (x-1).
Canceling this like term we get
![\frac{(4x-3)(x-1)} {(x+6)(x-1)}=\frac{4x-3} {x+6}](https://tex.z-dn.net/?f=%5Cfrac%7B%284x-3%29%28x-1%29%7D%20%7B%28x%2B6%29%28x-1%29%7D%3D%5Cfrac%7B4x-3%7D%20%7Bx%2B6%7D)
1 : .64 Since it is females to males you should divide the 9 males by 14 females to get 1 female equals x amount of males
Answer:
The answer is "
"
Step-by-step explanation:
The rectangle should also be symmetrical to it because of the symmetry to the y-axis The pole of the y-axis. Its lower two vertices are (-x,0). it means that
and (-x,0), and (x,0). Therefore the base measurement of the rectangle is 2x. The top vertices on the parabola are as follows:
The calculation of the height of the rectangle also is clearly 16-x^2, (-x,16,-x^2) and (x,16,-x^2).
The area of the rectangle:
![A(x)=(2x)(16-x^2)\\\\A(x)=32x-2x^3](https://tex.z-dn.net/?f=A%28x%29%3D%282x%29%2816-x%5E2%29%5C%5C%5C%5CA%28x%29%3D32x-2x%5E3)
The local extremes of this function are where the first derivative is 0:
![A'(x)=32-6x^2\\\\32-6x^2=0\\\\x= \pm\sqrt{\frac{32}{6}}\\\\x= \pm\frac{4\sqrt{3}}{3}\\\\](https://tex.z-dn.net/?f=A%27%28x%29%3D32-6x%5E2%5C%5C%5C%5C32-6x%5E2%3D0%5C%5C%5C%5Cx%3D%20%5Cpm%5Csqrt%7B%5Cfrac%7B32%7D%7B6%7D%7D%5C%5C%5C%5Cx%3D%20%5Cpm%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%5C%5C%5C%5C)
Simply ignore the negative root because we need a positive length calculation
It wants a maximum, this we want to see if the second derivative's profit at the end is negative.
![A''\frac{4\sqrt{3}}{3} = -12\frac{4\sqrt{3}}{3}](https://tex.z-dn.net/?f=A%27%27%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%20%3D%20-12%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%3C0%5C%5C%5C%5C2.%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%3D%20%5Cfrac%7B8%5Csqrt%7B3%7D%7D%7B3%7D%5C%5C%5C%5C%5Cvertical%20%5C%20dimension%5C%5C%5C%5C16-%28%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%29%5E2%3D%20%5Cfrac%7B32%7D%7B3%7D)