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nata0808 [166]
2 years ago
6

Can someone please help me find the horizontal asymptote of this graph I dont undestand

Mathematics
1 answer:
Andreas93 [3]2 years ago
7 0

9514 1404 393

Answer:

  p = 0

Step-by-step explanation:

The graph is of an exponential function. An exponential function always has a horizontal asymptote of 0.

  p(t) = 5×2^t

  p(t) → 0 as x → -∞

The horizontal asymptote is p = 0.

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6 students took a test, the dot plot shows the score of 5 of these students. The average score of students is 21 ? What was the
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4 0
2 years ago
What is the range of the data below 33, 38, 35.5,39.25, 31.75​
Lunna [17]

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uysha [10]
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2. 1-5(7)=-34 =/= 4
1-10=-9 =/= 1
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3. the slope is \frac{-10-(-5)}{-9-(-2)}= \frac{5}{7}. Answer C
8 0
2 years ago
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If function F is defined by f(x) = 3x^2 + 5, what is f(x-4)
shepuryov [24]

Answer:

The function f(x) when x=x-4 is f(x-4)=3x^2-24x+53  or f(x-4)=3x(x-8)+53

Step-by-step explanation:

Given function f is defined by f(x)=3x^2+5

Now to find the function f(x-4):

f(x)=3x^2+5

Put x=x-4 in the above function we get

f(x-4)=3(x-4)^2+5

f(x-4)=3(x^2-2(x)(4)+4^2)+5 (using the formula (a-b)^2=a^2-2ab+b^2)

f(x-4)=3(x^2-8x+16)+5

f(x-4)=3x^2-24x+48+5 (adding the like terms)

f(x-4)=3x^2-24x+53

f(x-4)=3x(x-8)+53

Therefore f(x-4)=3x^2-24x+53  or f(x-4)=3x(x-8)+53

The function f(x) when x=x-4 is f(x-4)=3x^2-24x+53  or f(x-4)=3x(x-8)+53

4 0
3 years ago
CAN ANYONE HELP ME WITH THIS QUESTION?
scoundrel [369]

Answer:

try 4

Step-by-step explanation:

8x4= 32

32-7=25

7 0
2 years ago
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