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3 years ago

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Walter wants computer software that cost $129. He has $57 saved. If he saves a maximum of $15 a week, the following inequality c

an be used to find how many weeks it will take Walter to save enough money to buy the software. 15x + 57 _> 129 (btw that’s an equal to symbol under it lol)
A. X_> 4.8
B. X_< 4.8
C. X_> 12.4
D. X_< 12.4

1 answer:

Lina20 [59]3 years ago

8 0

In this question, it is given that

Walter wants computer software that cost $129. He has $57 saved.

he saves a maximum of $15 a week, the following inequality can be used to find how many weeks it will take Walter to save enough money to buy the software.

$15x+57 \geq 129$

And we have to solve this inequality for x.

First we subtract 57 to both sides. And on doing so , we will get

$15x\geq 72$

Dividing both sides by 15.

$x \geq 4.8$ \

Correct option is A.

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PLZ HELP THIS IS TIMED!!!!

mihalych1998 [28]

Answer:

For the sequence is $-\frac{2}{3}$ ,-4 ,-24 ,-144 ,...

Hence the formula $f(x)=-\frac{2}{3}(6)^{x-1}$ for x=1,2,3,... represents the given geometric sequence

Step-by-step explanation:

Given sequence is $-\frac{2}{3}$ ,-4 ,-24 ,-144 ,...

To find the formula to describe the given sequence :

Let $a_1=\frac{-2}{3}$ , $a_2=-4$ , $a_3=-24$ ,...

First find the common ratio

$r=\frac{a_2}{a_1}$ here $a_1=\frac{-2}{3}$ and, $a_2=-4$

$=\frac{-4}{\frac{-2}{3}}$

$=\frac{4\times 3}{2}$

$=\frac{12}{2}$

$r=6$

$r=\frac{a_3}{a_2}$ here $a_2=-4$ and $a_3=-24$

$=\frac{-24}{-4}$

$=6$

$r=6$

Therefore the common ratio is 6

Therefore the given sequence is geometric sequence

The nth term of the geometric sequence is

$a_n=a_1r^{n-1}$

The formula which describes the given geometric sequence is

$f(x)=a_1r^{x-1}$ for x=1,2,3,...

$=\frac{-2}{3}6^{x-1}$ for x=1,2,3,...

Now verify that $f(x)=a_1r^{x-1}$ for x=1,2,3,... represents the given geometric sequence or not

put x=1 and the value of $a_1$ in $f(x)=a_1r^{x-1}$ for x=1,2,3,...

we get $f(1)=-\frac{2}{3}(6)^{1-1}$

$=-\frac{2}{3}(6)^0$

$=-\frac{2}{3}$

Therefore $f(1)=-\frac{2}{3}$

put x=2 we get $f(2)=-\frac{2}{3}(6)^{2-1}$

$=-\frac{2}{3}(6)^1$

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Therefore $f(2)=-4$

put x=3 we get $f(3)=-\frac{2}{3}(6)^{3-1}$

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Therefore $f(3)=-24$

Therefore the sequence is f(1),f(2),f(3),...

Therefore the sequence is $-\frac{2}{3}$ ,-4 ,-24 ,-144 ,...

Hence the formula $f(x)=a_1r^{x-1}$ for x=1,2,3,... represents the given geometric sequence is verified

Therefore the formula $f(x)=-\frac{2}{3}(6)^{x-1}$ for x=1,2,3,... represents the given geometric sequence

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