"In Grade 2 and early in Grade 3, students learned to use bar models to solve two-step problems involving addition and subtraction. This is extended in this chapter to include multiplication and division.
<span>Both multiplication and division are based on the concept of equal groups, or the part-part-whole concept, where each equal group is one part of the whole. In Grade 2, students showed this with one long bar (the whole) divided up into equal-sized parts, or units. This unitary bar model represents situations such as basket of apples being grouped equally into bags." </span>https://www.sophia.org/tutorials/math-in-focus-chapter-9-bar-modeling-with-multipli
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:
29m+37?
Step-by-step explanation:
<em>Pemdas</em>
5m+3(8m+7)+4^2
Parentheses first: 5m+24m+21+4^2
Exponents next: 5m+24m+21+16
No multiplication or division so addition is last:
29m+37
These two can't add together because one has a variable.
Answer:
Step-by-step explanation:
Given:
Area = 3x^3 - 16x^2 + 31x - 20
Base:
x^3 - 5x
Area of trapezoid, S = 1/2 × (A + B) × h
Using long division,
(2 × (3x^3 - 16x^2 + 31x - 20))/x^3 - 5x
= (6x^3 - 32x^2 + 62x - 40))/x^3 - 5x = 6 - (32x^2 - 92x + 40)/x^3 - 5x = 2S/Bh - Ah/Bh
= 2S/Bh - A/B
= (2S/B × 1/h) - A/B
Since, x^3 - 5x = B
Comparing the above,
A = 32x^2 - 92x + 40
2S/B = 6
Therefore, h = 1
His first payment is $100, thus a₁ = 100.
the next "term", month will be 1.1 times more than the one before, namely r = 1.1, the common ratio.


is the serie divergent or convergent?
well, to make it short, when the common ratio is 0 < | r | < 1, namely a fraction between 0 and 1, only then the serie is convergent, namely it reaches a fixed value, now in this case, 1.1 is a value larger than anything between 0 and 1, so no dice.