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FrozenT [24]
4 years ago
9

Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. f(x)

= 5x + 4 h(x) = x2 + 8x + 24
x = -3
x = 3
x = 4
x = 5
Mathematics
2 answers:
WINSTONCH [101]4 years ago
6 0
The exponential function is f(x)= 5^{x} +4
The polynomial function is h(x)= x^{2} +8x+24 

We want a value of x as such that f(x)\ \textgreater \ h(x), so

5^{x}+4 > x^{2} +8x+24

We are provided with four possible value of x. By process of trial and error, we will narrow down two values of x when the statement is changing from false to true

first, we try x=-3 by substituting into f(x) and h(x)

f(-3)=5^{-3}+4 = 4.008 and h(-3)= (-3)^{2}+8(-3)+24=9
so f(x)>h(x) which is not the inequality wanted

We try x=3 by substituting into f(x) and h(x)
f(3)= 5^{3}+4=129 and h(3)= (3)^{2}+8(3)+24=57, 
so f(x) > h(x) as wanted

We know that any values of x greater than three will give f(x) greater than h(x), so we narrow down to two values that make the statement changes from false to true.

There is one value between -3 and three that makes f(x) greater than h(x). From here we use the method of trial and error. 

We can start with the mid value between -3 and three which is 0
f(0)= 5^{0}+4=5
h(0)= 0^{2}+8(0)+24=24
f(x) is less than h(x) which isn't the inequality we want so here we narrow down the value of x must be between 0 and 3

We can try mid-value between 0 and three which is 1.5
f(1.5)= 5^{1.5}+4=15.18
h(1.5)= (1.5)^{2}+8(1.5)+24=38.25
f(x) is still less than h(x), so we now narrow down the value of x between 1.5 and 3

We try again using the mid-value between 1.5 and three which is 2.25
f(2.25)= 5^{2.25}+4=41.38
h(2.25)= 2.25^{2}+8(2.25)+24=47.0625
We still have f(x) less than h(x) so we can narrow down further the value of x between 2.25 and 3 (which is quite a short interval already)

We can keep trying by choosing a value of x says x=2.6
f(2.6)= 5^{2.6}+4=69.66
h(2.6)= 2.6^{2}+8(2.6)+24=51.56
We have f(x) greater then h(x) 

The narrowing down process continues where we now have the interval of x between 2.25 and 2.6. Keeping to one decimal place we can find the 
value of x=2.4 is the approximate value where f(x) is greater than h(x).

Answer: x=2.4
enyata [817]4 years ago
5 0

The correct answer is x = 5.

Trust me, I just did the quiz.

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Answer:

1. The ratio of their diameters is 8/5 = 8 : 5

2. The ratio of their surface area is (8/5)²

3. The ratio of their volume is (8/5)³

4.The area of the base of the larger cylinder is 128 cm²

Step-by-step explanation:

Given that the two cylinders are similar, we have;

Two cylinders are similar when the ratio of their heights is equal to the ratio of their radii

Therefore, we have;

1. The ratio of the height of the two cylinders = 8/5 = The radio of their radii = r₁/r₂

The ratio of their diameter = D₁/D₂ = 2·r₁/2·r₂ = r₁/r₂ = 8/5

The ratio of their diameters D₁/D₂ = 8/5 = 8 : 5

2. The surface area of the cylinders = 2·π·r·h + 2·π·r²

Therefore, we have;

(2·π·r₁·h₁ + 2·π·r₁²)/(2·π·r₂·h₂ + 2·π·r₂²) = (r₁·h₁ + r₁²)/(r₂·h₂ + r₂²)

h₁ = h₂ × 8/5

r₁ = r₂ × 8/5

= (8/5)²(r₂·h₂ + r₂²)/(r₂·h₂ + r₂²)  = (8/5)²

The ratio of their surface area = (8/5)²

3. The volume of the cylinder = π·r²·h

∴ The ratio of the volume = (π·r₁²·h₁)/(π·r₂²·h₂) = (8/5)³ × (π·r₂²·h₂)/(π·r₂²·h₂) = (8/5)³

The ratio of their volume = (8/5)³

4. The ratio of the area of the base of the larger cylinder to the area of the base of the smaller cylinder is (8/5)²

Therefore if the area of the base of the smaller cylinder is 50 cm², the area of the base of the larger cylinder = 50 cm² × (8/5)²  = 128 cm²

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How to write g(x)=2x+3/x+1 in the form of:g(x)=a/x+p +q
aliya0001 [1]
Hello,

Please write correctly.

g(x)= (2x+3)/(x+1) to be written like a/(x+p) + q

It is very simple!

(2x+3)/(x+1)= [(2x+2)+1]/(x+1)= (2(x+1)+1)/(x+1)= 2+1/(x+1)


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Answer:

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Simplify<br><br>-5/7(-(1 1/3-3/4)+1/3 / 4
koban [17]

Answer:

=5/21L+ -5/84

Step-by-step explanation:

=-(5/7) (-(1L/3-3/4)+1/3/4)

=(-5/7)(-(1L/3-3/4))+(-5/7)(1/3/4)

=5/21L= -15/28+ -5/84

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