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3 years ago

Use cross products to demonstrate whether the two ratios are equivalent in this problem: 4 over 6 times 14 over 21

1 answer:

3 0

4/6 = 14/21 When using cross product always multiply the left fractions numerator times the right side fractions denominator and the left side fractions denominator times the right side fractions numerator 4×21 = 6×14 84=84 They are equivalent because they both equal 84

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Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Anna71 [15]

Answer by JKismyhusbandbae: B) –10.6

Work/Explanation: Since the sequence slowly gets smaller, it is likely that each term is something added to (subtracted from) the previous term. To get from –7.9 to –8.8, it appears that the first has had –0.9 added to it. Continue adding –0.9 to get that the fourth term is –10.6.

8 0

3 years ago

use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 ), (15 comma

lyudmila [28]

Answer:

volume V of the solid

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.

Each slice is a quarter of cylinder 5/n thick and has a radius of

-k(5/n) + 5 for each k = 1,2,..., n (see picture)

So the volume of each slice is

$\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}$

for k=1,2,..., n

We then add up the volumes of all these slices

$\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}$

Notice that the last term of the sum vanishes. After making up the expression a little, we get

$\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2$

But

$\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)$

we also know that

$\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}$

and

$\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}$

so we have, after replacing and simplifying, the sum of the slices equals

$\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}$

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

$\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}$

and the volume V of our solid is

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

3 0

3 years ago

Mrs Richmond bought some apples and pears in the ratio 2:5. An apple costs 30 cents. A pear costs 15 cents mire than an apple. H

Lera25 [3.4K]

Answer:

$2.85

Step-by-step explanation:

30×2=60

30+15=45

60+(45×5)=?

45×5=225

60+225=285

$2.85

Cuz every 100cents is one dollar

5 0

3 years ago

If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n st ?

yulyashka [42]

Answer:

A) None

Step-by-step explanation:

1) $s^t$ shoudnt neccesarily be a factor of nst, for example, if s = 3, t = 4, and n = 12, then both s and t are factors of n, but $3^4 = 81$ is not a factor of nst = 144.

2) $(st)^2$ shoudnt neccesarily be a factor of nst. Let s be 4, let t be 6, and let n be 12. Then n is a factor of both s and t, but $(st)^2 = 24^2$ is not a factor of nst = 12*24. In fact, it is a greater number.

3) Again, s+t isnt necessarily a factor of nst, let s be 2 and t be 3. Then both s and t are factor of n = 12. However 5 = s+t is not a factor of nst = 72.

So, neither of the three options is guaranteed to be a factor of nst. In fact, for s = 4, t = 6, and n = 12, none of the three options are valid.

4 0

3 years ago

Find the area of a triangle formed by placing the vectors [3, 6] and [7, 1] tail-to-tail.

Usimov [2.4K]

Answer:

a. $A=10u^{2}$

b. View graph

c. 6.40u

Step-by-step explanation:

knowing that the triangle area is equal to base by heigh between two, then:

$A=\frac{bh}{2}=\frac{5*4}{2}=10u^{2}$

The length of the longest altitude of your triangle is:

$c^{2}=a^{2}+b^{2}=5^{2}+4^{2}=25+16=41\\ c=\sqrt{41}=6.40u$

finally it can be seen that the position of the triangle does not matter, as long as the base and heigh are maintained, the area of the triangle will be the same

6 0

3 years ago