Answer: 49/60
Step-by-step explanation:
(1/3+2/5)-1/4-(5/6-7/6)
Primero (5/6-7/6)
5/6-7/6=-2/6
1/3+2/5-1/4-(-2/6)
1/3+2/5-1/4+2/6
2/6=1/3
1/3+2/5-1/4+1/3
2/3+2/5-1/4
2/3=10/15
2/5=6/15
10/15+6/15-1/4
16/15-1/4
16/15=64/60
1/4=15/60
64/60-15/60=49/60
Answer:
Step-by-step explanation:
A, x=31
E, b=11
N, m=61
Y, b=15
I, a=80
G, k=-78
U, x=-13
R, u=88
O, y=13
A, w=104
H, d=-7
V, n=1/36
Sorry, don't want to give away all the answers!
Hope this helps!!!
Answer:

Step-by-step explanation:
From inspection of the graph, we can see that the curve intercepts the x-axis at (-4, 0), (-1, 0) and (3, 0)
Therefore,
x = -4 ⇒ x + 4 = 0
x = -1 ⇒ x + 1 = 0
x = 3 ⇒ x - 3 = 0
Because (-4, 0) touches the x-axis, then (x + 4)² will be a factor
So (x + 4)², (x + 1) and (x - 3) are all factors of the polynomial

If we multiply the constants, this will give us the y-intercept:
⇒ 4² x 1 x -3 = -48
From inspection of the graph, the y-intercept is -6
So to get from -48 to -6 we need to multiply -48 by 1/8
Therefore, n = 1/8

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.
The sequence is arithmetic since the common difference that is equal
d=11-15=7-11=-4
a(n) = a1 +(n-1)d
a1=15
=
a(n) = 15+(n-1)-4