16 times the quantity of numbers, in this case 5, which = 80, then subtract each number, 80-20-20-19-13=8. X should = 8
Answer:
Step-by-step explanation:
(i) Area of ABCD = (base1+base2)*height/2
= (6+12)*4/2 = 36 cm2
(ii) Volume = Area of ABCD * length = 36 * 25 = 900 cm3
(iii) Total SA = Area of ABCD *2 + 25*5*2 + 6*25 + 12*25
= 72 + 250 + 150 + 300
= 772cm2
<span>Determine the values of x on which the function f(x)=2x^2-x-15/4x^2-12x is discontinuous and verify the type of discontinuity at each point.
A.There is a vertical asymptote at 0 and a hole at 3.
B.There are vertical asymptotes at -5/2 & 0 and a hole at 3.
C.There is a vertical asymptote at 3 and a hole at 0.
D.There are vertical asymptotes at 0 & 3.
</span>
Result:
(-7/4)x^2-13x
The roots are:
x= -52/7
x=0
See attached pictures.
Answer:
![f(x)=4\sqrt[3]{16}^{2x}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D)
Step-by-step explanation:
We believe you're wanting to find a function with an equivalent base of ...
![4\sqrt[3]{4}\approx 6.3496](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B4%7D%5Capprox%206.3496)
The functions you're looking at seem to be ...
![f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x](https://tex.z-dn.net/?f=f%28x%29%3D2%5Csqrt%5B3%5D%7B16%7D%5Ex%5Capprox%202%5Ccdot2.5198%5Ex%5C%5C%5C%5Cf%28x%29%3D2%5Csqrt%5B3%5D%7B64%7D%5Ex%3D2%5Ccdot%204%5Ex%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D%5Capprox%204%5Ccdot%206.3496%5Ex%5C%20%5Cleftarrow%5Ctext%7B%20this%20one%7D%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B64%7D%5E%7B2x%7D%3D4%5Ccdot%2016%5Ex)
The third choice seems to be the one you're looking for.
Answer: each person would get 1.75 each
Step-by-step explanation:
devide 7.00 by 4