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3 years ago

How do i find XY??????

Mathematics

1 answer:

vladimir1956 [14]3 years ago

3 0

I think you have to 5 - 2.5

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Which linear inequality is represented by the graph?

julia-pushkina [17]

Please where is the graph

8 0

3 years ago

PLEASE EXTRA POINTS I REALLY NEED HELP!!! 1. A ski lift carries Carol up a slope at a rate of 5 mph.

gtnhenbr [62]

Hi sorry for the late response, I know the assignment was due at 5pm today but anyway here is the answer.

System:

d= 5(0.75-t)

d= 20t

20t = 3.75 -5t

Define Variables:

d = how far Carol Skied

t = time it took Carol to go up/down

Solve:

20t = 3.75 -5t

20t + 5t = 3.75 -5t + 5t

25t = 3.75

25t/25 = 3.75/25

t = 0.15

Now we plug in the time into the Distances:

20t = 3.75 - 5t

20(0.15) = 3.75 - 5(0.15)

3 = 3.75 - 0.75

3 = 3

Solution as Sentence:

Carol Skied 3 miles

4 0

3 years ago

Square root of 25/16 - square rot of 36/25

muminat

Square root of 25/16 is 5/4
square root of 36/25 is 6/5

square root of 25/16 - square root of 36/25 = 5/4 - 6/5 = (25 - 24)/20 = 1/20

8 0

3 years ago

M is the midpoint of YZ. If YM = x + 3, and YZ = 3x -1, find MZ

Elenna [48]

2(x+3)= 3x-1
2x+6=3x-1
X=7
Therefore
YM = 7+3= 10
And by definition of midpoint YM=MZ so MZ=10

3 0

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