Can Somebody Please Help With This?

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

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Answer:

Following is the conclusion and matching each tile with answer:

A. vertical stretch of a factor of 3 → g ( x ) = 3 (5)ˣ

B. vertical compression of a factor of 1/3 → g ( x ) = 1/3 (5)ˣ

C. horizontal stretch of a factor of 3 → $g(x) = 5^{\frac{1}{3}x}$

D. horizontal compression of a factor of 1/3 → $g(x) = 5^{3x}$

Step-by-step explanation:

let us suppose a function f ( x ), and a new function g ( x ) = a f ( x )

When a function is multiplied by a positive constant, let say 'a', you normally get a function the graph of which is vertically compressed or stretched with respect to the graph of the original function.

If the value of constant 'a' is greater than 1, the graph is vertically stretched.
If the value of constant 'a' is greater than 0 but lesser than 1, the graph will be vertically compressed.

So, the function g ( x ) = 1/3 (5)ˣ is compressed vertically by 1/3, as the multiplied value of constant 'a' = 1/3 is less than 1.

And, the function g ( x ) = 3 (5)ˣ is stretched vertically by 3, as the multiplied value of constant 'a' = 3 is greater than 1.

Now, let us suppose a function f ( x ), and a new function g ( x ) = f ( bx )

Observe the certain changes right inside the function. In other words, when the input of a function is multiplied by a positive constant, let say 'b', we get a function the graph of which is horizontally compressed or stretched.

If the value of constant 'b' is greater than 1, the graph is horizontally compressed by a factor of 1/b.
If the value of constant 'b' is greater than 0 but lesser than 1, the graph will be horizontally stretched 1/b.

So, the graph $g(x) = 5^{\frac{1}{3}x}$ is horizontally stretched by 3, as the value of constant 'b' is greater than 0 but lesser than 1.

And, the graph $g(x) = 5^{3x}$ is horizontally compressed by 1/3, as the value of constant 'b' is greater than 1.

Following is the conclusion and matching each tile with answer:

A. vertical stretch of a factor of 3 → g ( x ) = 3 (5)ˣ

B. vertical compression of a factor of 1/3 → g ( x ) = 1/3 (5)ˣ

C. horizontal stretch of a factor of 3 → $g(x) = 5^{\frac{1}{3}x}$

D. horizontal compression of a factor of 1/3 → $g(x) = 5^{3x}$

Keywords: vertical stretch, vertical compression, horizontal stretch, horizontal compression

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Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol

koban [17]

Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$