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Vesnalui [34]
4 years ago
8

After 3/4 of a minute he moved3/8 of a foot what is his speed per minute

Mathematics
1 answer:
denis23 [38]4 years ago
7 0
0,5 feet over minute
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The difference between -87 and -125.62?
marshall27 [118]

Answer:

38.62

Step-by-step explanation:

You can figure this out by converting these to a positive number, then subtracting the lower number from the higher number.

4 0
3 years ago
One thousand three hundred twenty feet are what part of a mile? Please give answer in lowest terms.
sashaice [31]
1320 ft out of a mile. 5280 feet in a mile so 

1320/5280 

keep simplifying until you get to 1/4
3 0
4 years ago
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What is the area of this polygon?<br><br> Enter your answer in the box.<br><br> units²
leva [86]

Answer:

45 sq. units

Step-by-step explanation:

We can draw a segment from F to S, splitting this figure into a rectangle and a triangle.

The area of a rectangle is given by the formula A=lw, where l is the length and w is the width.

The width of this figure is the distance of FW; this is 2.  The length of this figure is the distance of WC; this is 9.  This makes the area of the rectangle A=2(9) = 18 sq. units.

The area of a triangle is given by the formula A=1/2(b)(h), where b is the base and h is the height.

The base of the triangle is given by the distance of FS; this is 9.  The height of the triangle is given from N to segment FS; this is 6.  This makes the area of the triangle A=1/2(6)(9) = 27 sq. units.

This makes the total area 18+27 = 45 sq. units.

4 0
4 years ago
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- 20 = 2(n − 3) Asswer ?
jonny [76]

Answer:

Step-by-step explanation:

-20=2(n-3)

-20=2*n-2*3

-20=2n-6

-20+6-2n

-14/2=n

-7=n

4 0
3 years ago
Read 2 more answers
Use the Limit Comparison Test to determine whether the series converges.
lianna [129]

Answer:

The infinite series \displaystyle \sum\limits_{k = 1}^{\infty} \frac{8/k}{\sqrt{k + 7}} indeed converges.

Step-by-step explanation:

The limit comparison test for infinite series of positive terms compares the convergence of an infinite sequence (where all terms are greater than zero) to that of a similar-looking and better-known sequence (for example, a power series.)

For example, assume that it is known whether \displaystyle \sum\limits_{k = 1}^{\infty} b_k converges or not. Compute the following limit to study whether \displaystyle \sum\limits_{k = 1}^{\infty} a_k converges:

\displaystyle \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ $\leftarrow$ Series whose convergence is known\end{tabular}.

  • If that limit is a finite positive number, then the convergence of the these two series are supposed to be the same.
  • If that limit is equal to zero while a_k converges, then b_k is supposed to converge, as well.
  • If that limit approaches infinity while a_k does not converge, then b_k won't converge, either.

Let a_k denote each term of this infinite Rewrite the infinite sequence in this question:

\begin{aligned}a_k &= \frac{8/k}{\sqrt{k + 7}}\\ &= \frac{8}{k\cdot \sqrt{k + 7}} = \frac{8}{\sqrt{k^2\, (k + 7)}} = \frac{8}{\sqrt{k^3 + 7\, k^2}} \end{aligned}.

Compare that to the power series \displaystyle \sum\limits_{k = 1}^{\infty} b_k where \displaystyle b_k = \frac{1}{\sqrt{k^3}} = \frac{1}{k^{3/2}} = k^{-3/2}. Note that this

Verify that all terms of a_k are indeed greater than zero. Apply the limit comparison test:

\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ $\leftarrow$ Series whose convergence is known\end{tabular}\\ &= \lim\limits_{k \to \infty} \frac{\displaystyle \frac{8}{\sqrt{k^3 + 7\, k^2}}}{\displaystyle \frac{1}{{\sqrt{k^3}}}}\\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{k^3}{k^3 + 7\, k^2}}\right) = 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\end{aligned}.

Note, that both the square root function and fractions are continuous over all real numbers. Therefore, it is possible to move the limit inside these two functions. That is:

\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\\ &= \cdots \\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\\ &= 8\left(\sqrt{\frac{1}{\displaystyle 1 + \lim\limits_{k \to \infty} (7/k)}}\right) \\ &= 8\left(\sqrt{\frac{1}{1 + 0}}\right) \\ &= 8 \end{aligned}.

Because the limit of this ratio is a finite positive number, it can be concluded that the convergence of \displaystyle a_k &= \frac{8/k}{\sqrt{k + 7}} and \displaystyle b_k = \frac{1}{\sqrt{k^3}} are the same. Because the power series \displaystyle \sum\limits_{k = 1}^{\infty} b_k converges, (by the limit comparison test) the infinite series \displaystyle \sum\limits_{k = 1}^{\infty} a_k should also converge.

4 0
4 years ago
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