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Anettt [7]
3 years ago
14

A limited edition poster increases in value each year with an initial value of $18. After 1 year and an increase of 15% per year

, the poster is worth $20. 70. Which equation can be used to find the y value after x years
Mathematics
2 answers:
Gwar [14]3 years ago
7 0

Answer: The answer is y = 18(1.15)^x

Step-by-step explanation:

morpeh [17]3 years ago
3 0

Answer:

y = (18) * (1.15)^x

Step-by-step explanation:

A limited edition poster increases in value each year with an initial value of $18. After 1 year and an increase of 15% per year, the poster is worth $20. 70. Which equation can be used to find the y value after x years.

To find this equation, this is an exponential equation, meaning that the number increases at a rapid rate. In this case the post increases each year by 15% and started at $18. The equation you would use to find this is: y = a * b^x. We can fill in the a and b values based off the given information. Since the value is increasing by 15% we will add 1 to 15% to get 1.15. This will be the b value. The a value is our initial value which in this case is $18. Now we can plug everything in to get: y = (18) * (1.15)^x.

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Step-by-step explanation:

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Please help me with this ACT Prep problem.
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Answer:

Option C

Step-by-step explanation:

ΔABC and ΔDEF are similar triangles.

By the property of similar triangles, corresponding sides of the similar triangle are proportional.

\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}

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Using the definition of inverse (Definition 1, on Page 43) and nothing more, show that if A is an invertible matrix and c is a n
elena-s [515]

Answer:

The matrix cA is invertible and its inverse is \frac{1}{c}\cdot A^{-1}.

Step-by-step explanation:

Since the definition of the inverse matrix states that the inverse of matrix A is a matrix B such that:

A\cdot B=B\cdot A=I

we have to assume the form of such matrix. In our case we have the matrix cA, c\neq 0 and so, the constant c must be somehow eliminated from the equation. The most logical way to do so is to include \frac{1}{c} in the inverse. If we choose matrix B to be B=\frac{1}{c}\cdot A^{-1}, we will have this:

cA\cdot \frac{1}{c}\cdot A^{-1}=c\cdot \frac{1}{c}\cdot A\cdot A^{-1}=1\cdot I=I and

\frac{1}{c}\cdot A^{-1}\cdot cA=\frac{1}{c}\cdot c\cdot A^{-1}\cdot A=1\cdot I=I.

We can form the matrix B like this because we know from the text of the problem that the inverse matrix of A exists and that c is a nonzero number.

<u><em>Here is another way to solve this using the formula of the inverse matrix</em></u>

Since we know that the matrix A is invertible, it follows that its determinant is different from zero. Using the formula for the inverse matrix:

A^{-1}=\frac{1}{\det (A)}\cdot \text{Adj} (A)

we will assume the form of an inverse matrix of cA. We need to obtain the formula for the inverse of cA, so we first need to find \det (cA)\ \text{and}\ \text{Adj} (cA). Since the matrix cA is obtained from matrix A by multiplying every term with c, while calculating determinant we have a constant c that can be extracted from every column (or row) in front. Therefore, we have that

\det (cA)=c^n\cdot \det (A).

On the other hand, \text{Adj} (cA) consists of minors of the matrix cA. Therefore, when we extract the constant in front of such (n-1 \times n-1) determinants, we have c^{n-1} in each column (row). Including all this into the formula we have that:

(cA)^{-1}=\frac{1}{c^n\cdot \det (A)}\cdot c^{n-1} \text{Adj } (A)=\frac{1}{c\cdot \det (A)} \cdot \text{Adj} A=\frac{1}{c}\cdot A^{-1}.

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