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Igoryamba
3 years ago
10

What are two decimals that can be rounded to 1.5

Mathematics
1 answer:
irga5000 [103]3 years ago
4 0

Any 2 decimals?

In that case ==>  1.49753 and 1.46765

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Find the side lengths ! help pls
sergeinik [125]

(a) the given triangle is a isosceles triangle, therefore the two leg sides will be congruent, as well as the two base angles. It is given that one of the base angles ∠XYW is 70°, therefore, due to the law of a isosceles triangle, the measurement of ∠XWY is also 70°. Remember, a triangle's interior angles add up to 180°, so:

180 - (70 + 70) = 180 - 140 = 40

40° is your answer.

m∠X = 40°

(b) All sides are congruent, making it a equilateral triangle. If it is a equilateral triangle, then all the angles also have the same measurement. The total of the interior angle's measurement is 180°. Divide by the amount of angles, 3:

180/3 = 60

60° is your answer.

m∠V = 60°

~

7 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
There is a cuboids box with the
fenix001 [56]
Multiply the length by the width by the height.
8 0
3 years ago
Read 2 more answers
savannah has $5 in her wallet. she owes her brother 20 and owes her sister 18 . she earned 25 babysitting. write an algebraic ex
sladkih [1.3K]
An equation that represents this is

(5-20-18)+25

The steps to solve it are as follows

(-15-18)+25
-33+25= (-8)

she is still 8 dollars in debt
7 0
3 years ago
180 is 10 % more than witch number ?
Andrei [34K]
It is 10 percent more then 198.
6 0
2 years ago
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