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dimaraw [331]
3 years ago
6

Please answer all 3 questions! I provided them in pictures.

Mathematics
1 answer:
nirvana33 [79]3 years ago
5 0

Answer:

2) -3.1

3) -27/40

4) Step 2

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For f(x)=4x+1 and g(x)=x^2-5, find (g/f)(x)
ddd [48]

Answer:

(g/f)(x) =  \frac{g(x)}{f(x)}  =  \frac{ {x}^{2} - 5 }{4x + 1}

I hope I helped you^_^

3 0
3 years ago
Use properties of limits and algebraic methods to find the limit, if it exists. (If the limit is infinite, enter '[infinity]' or
soldi70 [24.7K]

Answer:

The limit of this function does not exist.

Step-by-step explanation:

\lim_{x \to 3} f(x)

f(x)=\left \{ {{9-3x} \quad if \>{x \>< \>3} \atop {x^{2}-x }\quad if \>{x\ \geq \>3 }} \right.

To find the limit of this function you always need to evaluate the one-sided limits. In mathematical language the limit exists if

\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) =L

and the limit does not exist if

\lim_{x \to a^{-}} f(x) \neq \lim_{x \to a^{+}} f(x)

Evaluate the one-sided limits.

The left-hand limit

\lim_{x \to 3^{-} } 9-3x= \lim_{x \to 3^{-} } 9-3*3=0

The right-hand limit

\lim_{x \to 3^{+} } x^{2} -x= \lim_{x \to 3^{+} } 3^{2}-3 =6

Because the limits are not the same the limit does not exist.

8 0
3 years ago
Can someone please help me with this?
dsp73

Answer:

S_{13} = 403

Step-by-step explanation:

the sum to n terms of an arithmetic series is

S_{n} = \frac{n}{2} (a + l) ← a is the first term and l the last term

here a = - 5 and l = 67 , then

S_{13} = \frac{13}{2} (- 5 + 67) = 6.5 × 62 = 403

3 0
2 years ago
Re-write the quadratic function below in Standard Form<br> y = -9(x - 1)(x + 2)
Rzqust [24]

Answer:xryuvb

Step-by-step explanation:

3 0
3 years ago
A shape is shown on the graph:
Setler79 [48]
Sorry it took me so long, the correct answer is B.
6 0
3 years ago
Read 2 more answers
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