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Strike441 [17]
2 years ago
10

F(x) = 3x + 1 and f-1 = f-1 = x-1 / 3 then f -'(7) =

Mathematics
1 answer:
Nutka1998 [239]2 years ago
7 0

Answer:

2

Step-by-step explanation:

f^-1 (7) = \frac{x-1}{3} = \frac{7-1}{3} = \frac{6}{3} = 2

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PLEASE HELP!!!
Aneli [31]
From the attached graphic, Half-life = ln (.5) / k

Half-life =.693147 / <span>0.1142
= </span><span><span><span>6.0695884413 </span> days

The value of "k" should be negative and should have units associated with it.
</span> </span>

3 0
3 years ago
Read 2 more answers
Find the remainder when 3x^3+ 6x-1 is divided by x-3​
Sauron [17]

Answer:

93

Step-by-step explanation:

by remainder theorem p(3) is the remainder

and therefore it is 93

3 0
3 years ago
What is the length of the diagonal of a 10 foot by 8 foot rectangle?
zloy xaker [14]
Diagonal of a rectangle formula:

Diagonal = √l² + w²
Diagonal = √10² + 8²
Diagonal = √100 + 64
Diagonal = √164
Diagonal = 12.806 feet.

Or you can use the pythagorean theorem. The diagonal of a rectangle is the hypotenuse of a triangle.

a² + b² = c² ; where a and b represents the lengths of the triangle and c is the hypothenuse. In this case, the lengths are 10 ft and 8ft.

10² + 8² = 100 + 64 = 164 ; This represents c².
c = √c² = √164 = 12.806 ft
6 0
3 years ago
Factorise completely:<br> 12a+8b
zavuch27 [327]
12a + 8b....common factor is 4
4(3a + 2b) <==
6 0
3 years ago
Solve using long division <br> Please
madreJ [45]

1. Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}

Steps:

\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}

\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}

\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}

\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}

\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}

2. Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}

Steps:

\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}

\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}

\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

4 0
3 years ago
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