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shepuryov [24]
3 years ago
12

How do I make a graph with y axis and x axis using numbers like 180

Mathematics
1 answer:
Rashid [163]3 years ago
8 0
Generally, you choose some convenient number for each small division of the graph to represent. Usually, 1, 2, 5, 10, and multiples of 10 times those are considered "convenient."

Here, it might do to use 5 units for each small division, so that 180 is 36 small divisions. If you have 10 small divisions per inch of your graph, then your graph will be about 4 inches high (if you start from zero).

Since the range of your graph is from 112 to 180, about 70 units, you could actually use 1 unit for each small division. Again, if you are using 10 divisions per inch, your graph will be about 7 inches high, so will fit nicely on a page. In this case, the lower edge will start at 110, not zero.

Since this is a temperature plot, you expect the curve to be exponential, with a lower limit of room temperature. Thus, you may want to use a graph with a range that includes room temperature at the low end. That will affect your choice of vertical scale somewhat.
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djyliett [7]

Answer:

C

Step-by-step explanation:

3 0
3 years ago
Karen earns $62.25 for working 9 hours. if the amount she earns varies with the number of hours she works how much will she earn
nata0808 [166]

Answer:

$103.75

Step-by-step explanation:

we know that

Karen earns $62.25 for working 9 hour

so

using proportion

Find out  how much she will  earn for working 15 hours

\frac{62.25}{9}\ \frac{\$}{h}=\frac{x}{15}\ \frac{\$}{h}\\\\x=62.25(15)/9\\\\x=\$103.75

6 0
3 years ago
(a) Suppose anxn has finite radius of convergence R and an ≥ 0 for all n. Show that if the series converges at R, then it also c
valina [46]

Answer:

a) See the proof below.

b) \sum \frac{(-x)^n}{n}

Step-by-step explanation:

Part a

For this case we assume that we have the following series \sum a)n x^n and this series has a finite radius of convergence R and we assume that a_n \geq 0 for all n, this information is given by the problem.

We assume that the series converges at the point x= R since w eknwo that converges, and since converges we can conclude that:

\sum a)n R^n < \infty

For this case we need to show that converges also for x=-R

So we need to proof that \sum a_n (-R)^n < \infty

We can do some algebra and we can rewrite the following expression like this:

\sum a_n (-R)^n = \sum (-1)^n a)n R^n and we see that the last series is alternating.

Since we know that \sum a_n x^n converges then the sequence {a_n R^n} must be positive and we need to have lim_{n\to \infty} a^n R^n = 0

And then by the alternating series test we can conclude that \sum a_n (-R)^n also converges. And then we conclude that the power series a_n x^n converges for x=-R ,and that complete the proof.

Part b

For this case we need to provide a series whose interval of convergence is exactly (-1,1]

And the best function for this \frac{(-x)^n}{n}

Because the series \sum \frac{(-x)^n}{n} converges to -ln(1+x) when |x| using the root test.

But by the properties of the natural log the series diverges at x=-1 because \sum \frac{1}{n} =\infty and for x=1 we know that converges since \sum \frac{-1}{n} is an alternating series that converges because the expression tends to 0.

6 0
3 years ago
Alexis bought 2 bags of apples for the fruit salad for her party. Each bag contained 13 apples. If she cuts each apple into 8 pi
Lerok [7]

Answer:

13*8=104

104*2=

<em>208</em>

Step-by-step explanation:

4 0
3 years ago
Find the value of x so that (-2, 4) is the midpoint between (x, 2) and (-5,3).
QveST [7]

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{x}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-5}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{-5+x}{2}~~,~~\cfrac{3+2}{2} \right)~~=~~\stackrel{midpoint}{(-2,4)}\implies \begin{cases} \cfrac{-5+x}{2}=-2\\[1em] -5+x=-4\\ \boxed{x=1} \end{cases}

6 0
3 years ago
Read 2 more answers
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