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Brrunno [24]
3 years ago
9

Ayudenme por favor nesecito ayuda con estas respuestas

Mathematics
1 answer:
malfutka [58]3 years ago
8 0

Answer:

ill help

Step-by-step explanation:

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If f(x)=2x−1, show that f(f(x))=4x−3. Find f(f(f(x))).
Allushta [10]

Answer: f(f(f(x))  =  8x - 7

Step-by-step explanation:

f(f(x)) means that we are evaluating f(x) with itself, then if f(x) = 2x - 1

f(f(x)) = 2*f(x) - 1 = 2*(2x - 1) - 1 = 4x - 2 - 1 = 4x - 3

f(f(x)) = 4x - 3

Then, we have f(f(f(x)) = 4*f(x) - 3

= 4*(2x - 1) - 3 = 8x - 4 - 3 =  8x - 7

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4 years ago
What is the slope of a line perpendicular to a line that contains the points (7, 1) and (-8 ,-3)
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Answer:

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Step-by-step explanation:

4 0
3 years ago
Find the product. (y^2)^5 · y^8<br>y^23 y^18 (y)^40<br>y^1/5
Klio2033 [76]
This question requires you to solve,

(y²)⁵×y⁸ =y⁽²ˣ⁵⁾ × y⁸
            = y¹⁰ × y⁸
            = y⁽¹⁰⁺⁸⁾
            = y¹⁸
The correct answer is y¹⁸
4 0
3 years ago
20 POINTS HURRY QUICK WILL GIVE BRAINLIEST
Mamont248 [21]

Answer:

A function is a rule that takes an input, does something to it,

and gives a unique corresponding output.

There is a special notation, called "function notation," that is used to represent this situation:

if the function name is  f , and the input name is  x ,

Step-by-step explanation:

4 0
3 years ago
A rock is dropped from a height of 100 feet. Calculate the time between when the rock was dropped and when it landed. If we choo
andre [41]

The time it takes between when the rock was dropped and when it landed is 1.33 seconds.

<h3>What is the time it takes for the rock to land?</h3>

The time it takes the rock to land on the ground from where it is dropped can be determined by using the function given.

The height of the function when it hits the ground will be zero, so the function can be written as:

\mathbf{h(t) =36t^2 -64}

\mathbf{0 =36t^2 -64}

64 = 36t²

\mathbf{t^2 = \dfrac{64}{36}}

t² =1.778

t = \sqrt{1.778}

t = 1.333 seconds

Learn more about calculating the time of a function here:

brainly.com/question/26898697

#SPJ1

6 0
2 years ago
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