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kakasveta [241]

3 years ago

12

Jose wants to rewrite 24+9 using the greatest common factor and the distributive property. Which expression should he write?

1 answer:

Reil [10]3 years ago

4 0

First get the factors of 24: 2*2*2*3Then the factors of 9: 3*3 Comparing the factors, the GCF is 3 Then you can rewrite the expression: 3*8 + 3*3 At this point, I'm not sure whether what you mean is really distributive property or not since this case is more of a factoring. 3*(8+3)

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emmainna [20.7K]

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5 0

3 years ago

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The exponential function f(x) = 3(5)x grows by a factor of 25 between x = 1 and x = 3. What factor does it grow by between x = 5

Delvig [45]

Answer:

B) 25

Step-by-step explanation:

Given exponential function:

$f(x)=3(5)^x$

The growth factor between $x=1$ and $x=3$ is 25.

To find the growth factor between $x=5$ and $x=7$

Solution:

The growth factor of an exponential function in the interval $x=a$ and $x=b$ is given by :

$G=\frac{f(b)}{f(a)}$

We can check this by plugging in the given points.

The growth factor between $x=1$ and $x=3$ would be calculated as:

$G=\frac{f(3)}{f(1)}$

$f(3)=3(5)^3$

$f(1)=3(5)^1$

Plugging in values.

$G=\frac{3(5)^3}{3(5)^1}$

$G=\frac{(5)^3}{(5)^1}$ (On canceling the common terms)

$G=(5)^{(3-1)}$ (Using quotient property of exponents $\frac{a^b}{a^c}=a^{(b-c)}$ )

$G=(5)^{2}$

∴ $G=25$

Similarly the growth factor between $x=5$ and $x=7$ would be:

$G=\frac{f(7)}{f(5)}$

$f(7)=3(5)^7$

$f(5)=3(5)^5$

Plugging in values.

$G=\frac{3(5)^7}{3(5)^5}$

$G=\frac{(5)^7}{(5)^5}$ (On canceling the common terms)

$G=(5)^{(7-5)}$ (Using quotient property of exponents $\frac{a^b}{a^c}=a^{(b-c)}$ )

$G=(5)^{2}$

∴ $G=25$

Thus, the growth factor remains the same which is =25.

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emmasim [6.3K]

Answer:

Hope this helps :) If it doesn't I can see what I can do

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3 years ago

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AnnyKZ [126]

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Answer:

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