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2 years ago

[[ CHECK THIS ASAP PLEASE ]] Is my answer correct or no?

Mathematics

1 answer:

Natalija [7]2 years ago

8 0

the shape is NOT a rhombus, because opposite sides are parallel but not all sides are congruent in length.

This shape is instead a parallelogram, where opposite sides are parallel but all sides do not have to be the same length

if 2 pairs of opposite sides are parallel, then opposite sides must be congruent, changing the angles will create nonparallel opposite sides that are unequal length and neither a parallelogram nor rhombus

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there is 4/5 lb of sand in science supplies. if one scoop of sand weighs 1/20 of a lb how many scoops of sand can maria get from

tatuchka [14]

We want to divide whole sand into scoops, so we have to divide 4/5 by 1/20. Lets do this

$\frac{ \frac{4}{5} }{ \frac{1}{20} }$
We know that if we want to divide 2 fraction we can multiply first by inverse of second fraction instead of dividing.,so:
$\frac{ \frac{4}{5} }{ \frac{1}{20} }$
$\frac{4}{5}* \frac{20}{1}= \frac{4*20}{5*1} = \frac{80}{5}= 16$ scoops - its our result.

5 0

2 years ago

Space Shuttle Marion is observing the launch of a space

Vera_Pavlovna [14]

Using the slope concept, it is found that the shuttle traveled 3.2 miles while Marion was watching.

<h3>What is a slope?</h3>

The slope is given by the <u>vertical change divided by the horizontal change</u>, and it's also the tangent of the angle of depression.

Initially, the angle of elevation is of 16º, with an elevation of 1 mi, hence:

$\tan{16^\circ} = \frac{1}{x_1}$

$x_1 = \frac{1}{\tan{16^\circ}}$

$x_1 = 3.5$

Then, at the end, the angle will be of 74º, with the elevation remaining constant, hence:

$\tan{74^\circ} = \frac{1}{x_2}$

$x_2 = \frac{1}{\tan{74^\circ}}$

$x_2 = 0.3$

The distance is:

$d = x_1 - x_2 = 3.5 - 0.3 = 3.2$

The shuttle traveled 3.2 miles while Marion was watching.

More can be learned about the slope concept at brainly.com/question/18090623

3 0

2 years ago

The graph shows a journey in a car Help Fast!!!

Sever21 [200]

Answer: i think the answer would be c

i hope this helps you out!

Step-by-step explanation:

3 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)

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