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Elenna [48]
2 years ago
7

you are allowed to work a total of no more that 30 hours each week at your two jobs. lawn mowing pays $5 per hour and babysittin

g pays $8 per hour.you need to earn at least $300 per week.You decide to write a system of inequalities to help determine the amount of time you can work at each job, where x represents the number of you mow lawns and y represents the number of hours you babysit.You write the first inequality you would use for this system.
Mathematics
1 answer:
Sindrei [870]2 years ago
8 0

Answer:

you can't logically do this, even if you spend the entire time babysitting, you will not have enough for the 300

Step-by-step explanation:

Sorry this isn't an answer

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Antoine and Adriane are each at the top of a Ferris wheelAfter one revolution each will again be at the topWhich sinusoid (sine
Valentin [98]

Answer(s):

\displaystyle y = 100sin\:(\frac{\pi}{50}x + \frac{\pi}{2}) + 100 \\ y = 100cos\:\frac{\pi}{50}x + 100

Step-by-step explanation:

\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 100 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-25} \hookrightarrow \frac{-\frac{\pi}{2}}{\frac{\pi}{50}} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{100} \hookrightarrow \frac{2}{\frac{\pi}{50}}\pi \\ Amplitude \hookrightarrow 100

<em>OR</em>

\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 100 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{100} \hookrightarrow \frac{2}{\frac{\pi}{50}}\pi \\ Amplitude \hookrightarrow 100

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of <em>sine</em>, then by all means, go for it, but be careful and follow what is explained here. Now, as you can see, the photograph on the right displays the trigonometric graph of \displaystyle y = 100sin\:\frac{\pi}{50}x + 100, in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the <em>co</em><em>sine</em> graph [photograph on the left], accourding to the <u>horisontal shift formula above</u>. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the <em>sine</em> graph [photograph on the right] is shifted \displaystyle 25\:unitsto the right, which means that in order to match the <em>cosine</em> graph [photograph on the left], we need to shift the graph BACKWARD \displaystyle 25\:units,which means the C-term will be negative, and by perfourming your calculations, you will arrive at \displaystyle \boxed{-25} = \frac{-\frac{\pi}{2}}{\frac{\pi}{50}}.So, the sine graph of the cosine graph, accourding to the horisontal shift, is \displaystyle y = 100sin\:(\frac{\pi}{50}x + \frac{\pi}{2}) + 100.Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits \displaystyle [220, 100],from there to \displaystyle [120, 100],they are obviously \displaystyle 100\:unitsapart, telling you that the period of the graph is \displaystyle 100.Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the <em>midline</em>. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at \displaystyle y = 100,in which each crest is extended <em>one hundred units</em> beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

**I knew exactly what you were talking about the moment you posted this, so here was what you were looking for. It really does not matter which sinusoid and person you select because you will get the same information either way.

I am delighted to assist you at any time.

6 0
2 years ago
Please help find domain and range
Oxana [17]
<h3>Answer:  Choice A</h3>
  • Domain:  x > 4
  • Range: y > 0

========================================================

Explanation:

We want to avoid having a negative number under the square root. Solving x-4 \ge 0 leads to x \ge 4

So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.

So in reality, the domain is x > 4.

-------------

The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.

We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression \frac{3}{\sqrt{x-4}} is positive because positive/positive = positive.

Therefore, the range of the given equation is y > 0

-------------

The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).

3 0
2 years ago
Read 2 more answers
9.50 One way to evaluate the effectiveness of a teaching assistant is to examine the scores achieved by his or her students on a
EleoNora [17]

Answer:

Step-by-step explanation:

Given that a professor sets a standard examination at the end of each semester for all sections of a course. The variance of the scores on this test is typically very close to 300.

H_0: s^2 = 300\\H_a: s^2 \neq 300

(Two tailed test for variance )

Sample variance =480

We can use chi square test for testing of hypothesis

Test statistic = \frac{(n-1)s^2}{\sigma^2} \\=62.4

p value = 0.0100

Since p <0.05 our significance level, we reject H0.

The sample variance cannot be claimed as equal to 300.

5 0
3 years ago
7p²-30p + 8 how do you factor this trinomial
tamaranim1 [39]

Step-by-step explanation:

7p²-30p+8

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3 years ago
If you roll a fair die 100 times and 35 of the rolls result in a two, what would be the probability of the outcome of a 2,
EastWind [94]

Answer:

56

Step-by-step explanation:

5 0
2 years ago
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