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3 years ago

A rectangle has a length of 12 meters and a width of 400 centimeters. What is the perimeter,in cm, of the rectangle?

Mathematics

2 answers:

harina [27]3 years ago

4 0

Answer:

824cm

Step-by-step explanation:

8090 [49]3 years ago

4 0

Answer:

hcsdysdjcby

Step-by-step explanation:

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If y varies inversely as x and y=24 when x=8, find y when x is 4

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Y would equal 12 if x were to equal 4. You can think of it as a proportion.

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3 years ago

2 sides of a triangle are congruent. The base length is unknown. The top angle is 40 degrees and the bottom left angle is x degr

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Step-by-step explanation:

both sides of the triangle are 70 because they equal 140 plus 40 is 180

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Answer:

Distributive Property

Step-by-step explanation:

7 equals (3+4).

Therefore, 4*7 equals 4*(3+4).

By distributive property, this will equal (4*3) + (4*4)

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3 years ago

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3 years ago

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

4 0

3 years ago