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3 years ago

10

Jamie jogged a total distance of 6 and 1 over 2 miles during the months of March and April. If Jamie only jogged 1 over 6 mile e

very day, which expression shows the number of days she went jogging?

2 answers:

jasenka [17]3 years ago

5 0

6 1/2 divided by 1/6

Rashid [163]3 years ago

3 0

Answer:

6 1/2 divided by 1/6

Step-by-step explanation:

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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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X = 9
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2 years ago

2x+3=4 3x+y=solve by graphing and plz show work

ch4aika [34]

Answer:

2x+3=4

-3 -3

2x=1

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Maya wants to know how many games the seventh grade students at her school play on their phones. Which is the BEST population fo

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The best population for Maya's school would be all the seventh grade students at Maya's school, because that is who she wants to know how many games they play.

D. All the seventh grade students at Maya's school.

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