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schepotkina [342]
3 years ago
14

1. Find an algebraic representation for an exponential function if it is known that f(x + 1) = 4.f(x)

Mathematics
1 answer:
MrMuchimi3 years ago
4 0

f(x)=4^{x-1} is the algebraic representation for an exponential function

Step-by-step explanation:

Given:

f(x + 1) = 4.f(x)

f(3) = 16

To Find:

Algebraic representation for an exponential function=?

Solution:

From the formula f(x+n) = 4^n f(x)

when n= 1, x= 3

f(3+1)= 4(1)f(3)

f(4)= 4f(3)

Substituting the value of f(3)

f(4)= 4f(3)

f(4)= 4 x 16  

f(4)= 64

f(4)=4^2

f (5) = 4^2 x 16

f (5) = 4^2 x  4^2

f(5)= 4^4

Similarly,

F(6) = 4^5

Hence, f(x)=4^{x-1}

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Now, we solve for X:

\Rightarrow X=\frac{\sin(60)}{\sin(30)}\Rightarrow X=\sqrt[]{3}

So, the length of X is sqrt(3).

5 0
1 year ago
What is the value of t? <br> 2<br> 3<br> 6<br> 8
shepuryov [24]

Answer:

3

Step-by-step explanation:

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3 years ago
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Y = 7x is that a direct variation or no
natta225 [31]

Answer:

Step-by-step explanation:

yes i think so

5 0
3 years ago
a pentagon is a regular pentagon that has 5 sides, so each side has the same length that is (x - 1.5)cm. The perimeter is 22.5 c
Ivanshal [37]

Answer:

Step-by-step explanation:

5(x - 1.5) = 22.5

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8 0
2 years ago
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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