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hoa [83]
3 years ago
5

Does anyone know the answer pleaseeee

Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
8 0

Answer:

A, B, C are false. D, E, F are true

Step-by-step explanation:

let's use the number 3 as a case study

A. 3²-1 = 9 - 1 = 8 (false)

B. 3(3-1) = 3 × 2 = 6 (false)

C. (3-1)² = 2² = 4 (false)

D. 3² + 2 = 9 + 2 = 11 (true)

E. 3(3-2) = 3 × 1 = 3 (true)

F. (3-2)² = 1² = 1 (true)

Sav [38]3 years ago
7 0
I’m pretty sure a, d, e and f are correct :)
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4 0
4 years ago
Equivalate a +4 when a=7
joja [24]

Answer:

11

Step-by-step explanation:

Plug in the variable into the equation. 7+4=11

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LuckyWell [14K]
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2 years ago
The line represented by y = 3x − 2 and a line perpendicular to it intersect at R(3, 7). What is the equation of the perpendicula
laiz [17]

Answer:

y= -1/3(x) + 8

Step-by-step explanation:

slope of the 1st line = 3

slope of the line perpendicular to the first is its opposite reciprocal, so

slope of the 2nd line = -1/3

now we have a point and the slope of the perpendicular line, so we use the point slope formula to get:

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7 0
3 years ago
Let f(x) = (x − 3)−2. find all values of c in (1, 4) such that f(4) − f(1) = f '(c)(4 − 1). (enter your answers as a comma-separ
Verizon [17]

If the function is (x-3)^{-2} and  f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.

Given function is (x-3)^{-2} and  f(4) − f(1) = f '(c)(4 − 1).

In this question we have to apply the mean value theorem, which says that given a secant line between points A and B, there is at least a point C that belongs to the curve and the derivative of that curve exists.

We begin by calculating f(2) and f(5):

f(2)=(2-3)^{-2}

f(2)=1

f(5)=(5-3)^{-2}

f(5)=1

And the slope of the function:

f^{1}(x)=f(5)-f(2)/(5-2)

f^{1}(c)=0

Now,

f^{1} (x)=-2*(x-3)^{-3}

=-2(x-3)^{-3}

=0

-2 is not equal to 0. So there is not any answer.

Hence if the function is (x-3)^{-2} and  f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.

Learn more about derivatives at brainly.com/question/23819325

#SPJ4

7 0
2 years ago
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