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Alex
3 years ago
13

Can someone please help me with my homework and explain it to me I will mark as brainlyist

Mathematics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

121.76

Step-by-step explanation:

the original price is 83.97. you would have to find 45% of that. you would do this by putting x/83.97= 45/100

multiply 45 and 83.97 to get 3778.65. then divide that by 100 to get 37.7865. Add that to 83.97 to get 121.7565. round it to the nearest hundreth to get the total of 121.76 dollars

mark as brainlyest and answer my question

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Domain of the function of a graph with coordinates (-3,-2)
olga2289 [7]

Answer:

Domain is -3

Step-by-step explanation:

Given that coordinate of a graph is (-3, -2)

We have to find out the domain.

We know that if A and B are two sets, a mapping from A to B is the subset of cartesian product AxB.

Domain is the set of values of A which have images in B.

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Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

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1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

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2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

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3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

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4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

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5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

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