Simplify the expression?

Colleen plans to start a small business by making and selling skirts. A sewing machine costs $250, and she estimates that she wi

Dmitrij [34]

Let C(x) be the function which calculates the total cost of making x skirts.

There is going to be a fixed constant, 250, to which we will add 15$ per x skirts.

So C(x)=250+15x

Answer: 250+15x

7 0

3 years ago

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Find the value of (-4)(6)(-7) -178 168 178 -168

kakasveta [241]

Answer:

168

Explanation:

Before we begin, remember the following:

+ve * +ve = +ve -ve * -ve = +ve

+ve * -ve = -ve -ve * +ve = +ve

Now, for the given problem we have:

(-4) * (6) * (-7)

Let's take the first two terms:

(-4) * (6)

Based on the above rules, the product will be negative

Therefore,

(-4) * (6) = -24

Now, the expression became:

(-24) * (-7)

Again, based on the above rule, the product here will be positive

Therefore,

(-24) * (-7) = 168

Hope this helps :)

4 0

3 years ago

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The 4 in 400,000 has a value of 400 yes or no

Oduvanchick [21]

I'm pretty sure its not.

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3 years ago

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What you use an area model how can you tell if two fractions are not equivalent

svp [43]

Answer:

multiply or simplify

Step-by-step explanation:

8 0

3 years ago

For what values of x is log base 0.8 (x+4)>log base 0.4 (x+4) Thank you!

Radda [10]

We are seeking the solution of the inequality:

$\displaystyle{ \log_{0.8}(x+4)\ \textgreater \ \log_{0.4}(x+4)$ .

We recall that a log function $f(x)=\log_b(x)$ is either increasing or decreasing:

i) it is increasing if b>1,

ii) it is decreasing if 0<b<1.

Consider the functions $\displaystyle{ \log_{0.8}(x)$ and $\displaystyle{ \log_{0.4}(x)$ .

The graphs of these functions both meet at x=1 (clearly), and after 1 they are both negative. So from 0 to 1 one of them is larger for all x, and from 1 to infinity the other is larger. (Being strictly decreasing, their graphs can only intersect once.)

We can check for a certain convenient point, for example x=0.8:

$\displaystyle{ \log_{0.8}(0.8)=1$ and

$\displaystyle{ \log_{0.4}(0.8)=\log_{0.4}(0.4\cdot 2)=\log_{0.4}(0.4)+\log_{0.4}(\cdot 2)=1+\log_{0.4}(2)$ .

Now, $\displaystyle{ \log_{0.4}(2)$ is negative since we already explained that for x>1 both functions were negative. This means that

$\displaystyle{ \log_{0.8}(0.8)\ \textgreater \ \log_{0.4}(0.8)$ , and since $0.8\in (0, 1)$ , then this is the interval where $\displaystyle{ \log_{0.8}(x)\ \textgreater \ \log_{0.4}(x)$ .

So, now considering the functions $\displaystyle{ \log_{0.8}(x+4)$ and $\displaystyle{ \log_{0.4}(x+4)$ , we see that

x+4 must be in the interval (0,1), so we solve:

0<x+4<1, which yields -4<x<-3 after we subtract by 4.

Answer: (-4, -3). Attached is the graph generated using Desmos.

7 0

3 years ago