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gulaghasi [49]
3 years ago
7

Evaluate the following expression if w = 9, x = 7, and y = 1. 11y - 5 106 16 6 7

Mathematics
2 answers:
adell [148]3 years ago
5 0

Answer:

6

Step-by-step explanation:

Luckily for us, the expression only contains a variable y, so we can disregard the values for variables x and w.

Note, that y = 1.

With that in mind, let's evaluate the expression by plugging the value of y into the expression.

11y - 5 = ?

11(1) - 5 = ?

11 - 5 = ?

6 = ?

Hence, our expression evaluated to 6 when y = 1.

Cheers.

irga5000 [103]3 years ago
3 0

Answer:

6

Step-by-step explanation:

We are given the expression:

11y-5

and asked to evaluate when w=9, x=7, and y=1.

The variables w and x are not included in the expression, so we only have to focus on the variable y.

y is equal to 1, so we can plug 1 in for y.

11y-5

11(1)-5

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition and Subtraction.

First, multiply 11 and 1.

11*1=11

11-5

Next, subtract 5 from 11.

6

The expression 11y-5 when evaluated for y=1 is equal to 6.

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Answer:

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Step-by-step explanation:

There are a few ways you can write the equivalent of this.

1) Distribute the minus sign. The starting numerator is -(u-6). After you distribute the minus sign, you get -u+6. You can leave it like that, so that your equivalent form is ...

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Or, you can rearrange the terms so the leading coefficient is positive:

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2) You can perform the division and express the result as a quotient and a remainder. Once again, you can choose to make the leading coefficient positive or not.

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or

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Of course, anywhere along the chain of equal signs the expressions are equivalent.

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3) You can separate the numerator terms, expressing each over the denominator:

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4) You can also multiply numerator and denominator by some constant, say 3:

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You could do the same thing with a variable, as long as you restrict the variable to be non-zero. Or, you could use a non-zero expression, such as 1+x^2:

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Answer

Exponential function is in the form of : y =a(1+r)^x .....[1]

where a is the initial amount and r is the growth rate and (1+r) is the growth factor.

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On comparing with equation [1] we have;

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