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dedylja [7]
2 years ago
10

Jessica is putting money into a checking account. Let y represent the total amount of money in the account (in dollars). Let x r

epresent the number of weeks Jessica has been adding money. Suppose that x and y are related by the equation =+55030xy.
Mathematics
1 answer:
uranmaximum [27]2 years ago
6 0

After differentiating, the amount when x=0 yields starting amount,

300 dollars.

What is diffrentiation?

  • A technique for determining a function's derivative is differentiation.
  • A technique for determining a function's derivative is differentiation. Mathematicians use a procedure called differentiation to determine a function's instantaneous rate of variation based on a particular variables. The most typical illustration is velocity, which is the rate at which a distance changes in relation to time.
  • In mathematics, differentiation is used to determine change rates. 

Differentiating the given equation with respect to x,

Given equation,

y= 55x+ 300

\frac{dy}{dx} =\frac{d(55x+300)}{dx} = 55+0 =$55 dollars.

The amount when x=0 yields starting amount,

y=55(0)+300=300

Learn more about differentiation here:

brainly.com/question/25081524

#SPJ9

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3 years ago
In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.
DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2

                           + 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5

5C_0(\frac{1}{ax})^5(2ax^2)^0  =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1  = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\

5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

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3 years ago
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