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laila [671]
3 years ago
14

Susan enlarged a rectangle with a height of 6 cm and length of 13 cm on her computer. The length of the new rectangle is 19.5 cm

. Find the height of the new rectangle.
Mathematics
1 answer:
Karolina [17]3 years ago
4 0
We have that enlarging the rectangle in a computer setting keeps the ratio of the lines constant. Hence, height over length is the same. For the first rectangle this is 6/13. For the second one, if x is the new height, this is x/19.5.
Hence x/19.5=6/13. Multiplying both parts by 19.5 we have that x=19.5*6/13=9. Thus, the new height is 6 cm.
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Verify the identity
Thepotemich [5.8K]

<u>Given</u>:

Given that we need to prove the identity \sin x+\sin x \tan ^{2} x=\tan x \sec x

<u>Proof</u>:

Step 1: Factor out the common term sin x, we get;

\sin x\left(1+\tan ^{2} x\right)=tan \ x \ sec \ x

Step 2: Using the identity 1+tan^2 x=sec^2x

\sin x \sec ^{2} x=tan \ x \ sec \ x

Step 3: Reciprocating sec x, we get;

\sin x \cdot \frac{1}{\cos ^{2} x}=tan \ x \ sec \ x

Step 4: Splitting the denominator, we have;

\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}=tan \ x \ sec \ x

Simplifying, we get;

\tan x \sec x=\tan x \sec x

Thus, the identity is proved.

5 0
3 years ago
jem pastes 21 photos and 15 postcards in a scrap album. she puts 3 items on each page. how many pages does jem fill in the scrap
ddd [48]
12 pages

Each side count as a page

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6 0
3 years ago
A bicycle tire has a radius of 11 inches. To the nearest inch, how far does the tire travel when it makes 7 revolutions?
jok3333 [9.3K]

Answer:

d = 483.56 inches

Step-by-step explanation:

We have,

Radius of a bicycle tire is 11 inches. It is required to find the distance traveled by the distance covered by the tire when it makes 7 revolutions. One revolution means the circumference of circle. So,

C=2\pi r\\\\C=2\times 3.14\times 11\\\\C=69.08\ \text{inches}

In 7 revolution, it will cover a distance of 7C i.e.

C=69.08\times 7\\\\C=483.56\ \text{inches}

So, the tire will cover a distance of 483.56 inches in 7 revolutions

8 0
3 years ago
A) Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given cur
Leno4ka [110]

(a) See the attached sketch. Each shell will have a radius <em>y</em> chosen from the interval [2, 4], a height of <em>x</em> = 2/<em>y</em>, and thickness ∆<em>y</em>. For infinitely many shells, we have ∆<em>y</em> converging to 0, and each super-thin shell contributes an infinitesimal volume of

2<em>π</em> (radius)² (height) = 4<em>πy</em>

Then the volume of the solid is obtained by integrating over [2, 4]:

\displaystyle 4\pi \int_2^4 y\,\mathrm dy = 2\pi y^2\bigg|_{y=2}^{y=4} = 2\pi (4^2-2^2) = \boxed{24\pi}

(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - <em>x</em> (this is the distance between a given <em>x</em> value in the orange shaded region to the axis of revolution) and a height of 8 - <em>x</em> ³ (and this is the distance between the line <em>y</em> = 8 and the curve <em>y</em> = <em>x</em> ³). Then each shell has a volume of

2<em>π</em> (9 - <em>x</em>)² (8 - <em>x</em> ³) = 2<em>π</em> (648 - 144<em>x</em> + 8<em>x</em> ² - 81<em>x</em> ³ + 18<em>x</em> ⁴ - <em>x</em> ⁵)

so that the overall volume of the solid would be

\displaystyle 2\pi \int_0^2 (648-144x+8x^2-81x^3+18x^4-x^5)\,\mathrm dx = \boxed{\frac{24296\pi}{15}}

I leave the details of integrating to you.

3 0
2 years ago
If m AC =86°, find m &lt;AVC
bulgar [2K]
M<AVC = 43 degrees, because it is the angle's measure is 1/2 of the arc's
7 0
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