Let a and b be positive integers. 23^a x 23^b

Mathematics

2 answers:

Readme [11.4K]3 years ago

7 0

Answer:

23^(a+b)

Step-by-step explanation:

A^x*A^y = A^(x+y)

you can think of it like this

A^x = A*A*A*.......*A x times

A^y = A*A*A*....*A y times

therefore A^x * A^y = A*A*A*...*A (x+y) times

=A^(x+y),

so in our case A = 23, and x and y were a and b, so 23^a * 23^b = 23^(a+b)

Ket [755]3 years ago

6 0

Answer:

23 ^ ( a+b)

Step-by-step explanation:

23^a *23^b

Since the bases are the same, we can add the exponents

23 ^ ( a+b)

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5 (x+9) = 3 (x +8) + 2x

zmey [24]

Answer:

No solution

Step-by-step explanation:

Distribute

5x+45 = 3x+24+ 2x

Combine terms

45=5x-5x

45=0

This statement is not true so it is no solution

8 0

2 years ago

Write the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.

Gnoma [55]

Answer:

$\frac{A}{(x-1)} + \frac{B}{(x-9)}$

Step-by-step explanation:

Given the expression $\dfrac{1}{(x-1)(x-9)}$ , we are to write the expression as a partial fraction. Writing as a partial fraction means rewriting the expression a s a sum of two or more expression.

Before we will do this we will need to check the nature of the function at the denominator whether it is linear, quadratic or a repeated function. According to the question, the denominator at the denominator is a linear function and since it is a linear function, we can separate both linear function without restriction as shown;

$\dfrac{1}{(x-1)(x-9)} = \frac{A}{(x-1)} + \frac{B}{(x-9)}$ where A and B are the unknown constant which are numerical values.