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3 years ago

Help me on 16 and 17 please as fast as possible it’s due tomorrow

Mathematics

1 answer:

mihalych1998 [28]3 years ago

7 0

Answer:

Step-by-step explanation:

16.) 2/3

You add up the fractions and then reduce.

(5/12 + 3/12 = 8/12 = 2/3)

17.) No because if you add all the fractions together, you only get 7/8. This means the project is not complete yet.

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WILL GIVE BRAINLIEST What are the like terms in the algebraic expression? Negative a squared b + 6 a b minus 8 + 5 a squared b m

Minchanka [31]

Answer: B

Negative a squared b and 5 a squared b

Step-by-step explanation:

Given that:

Negative a squared b + 6 a b minus 8 + 5 a squared b minus 6 a minus b. That is,

- a^2b + 6ab - 8 + 5a^b - 6a - b

Collecting the like term by rearranging the expression

5a^2b - a^2b + 6ab - 6a - b

The like terms in the expression above are

5a^2b - a^2b.

The correct option is B:

Negative a squared b and 5 a squared b or (-a^2b and 5a^b)

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3 years ago

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Explain how to identify the terms of 3y - 4 - 5y

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Well the answer is -2y-4 but the different terms in this problem are 4 and 5y and 3y where 5y and 3y are like terms

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2 years ago

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Natasha2012 [34]

Answer:

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Step-by-step explanation:

The surface has a width of 11 meters and a length of 13 meters.

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2 years ago

1) Find the slope and the y-byintercept of the line that passes through the points (1,5) and (3,9)

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Answer: slope= 2, y-intercept= 3

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3 years ago

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KIM [24]

Answer:

(a) 0

(b) f(x) = g(x)

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{x^2+2x+1}{x^2-1}$

Part (a)

Factor the numerator and denominator of the given rational function:

$\begin{aligned} \implies f(x) & = \dfrac{x^2+2x+1}{x^2-1} \\\\& = \dfrac{(x+1)^2}{(x+1)(x-1)}\\\\& = \dfrac{x+1}{x-1}\end{aligned}$

Substitute x = -1 to find the limit:

$\displaystyle \lim_{x \to -1}f(x)=\dfrac{-1+1}{-1-1}=\dfrac{0}{-2}=0$

Therefore:

$\displaystyle \lim_{x \to -1}f(x)=0$

Part (b)

From part (a), we can see that the simplified function f(x) is the same as the given function g(x). Therefore, f(x) = g(x).

Part (c)

As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero). Therefore, the quotient approaches infinity.

$\displaystyle \lim_{x \to 1^+} f(x)=\dfrac{\to 2^+}{\to 0^+}=\infty$

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