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rodikova [14]
3 years ago
10

A carnival ride has a sign at the beginning of the line that says "Must be at least 48 inches tall to ride." Which of the follow

ing inequalities best represents the allowable height of riders?
h > 48
h < 48
h ≥ 48
h ≤ 48
Mathematics
2 answers:
Korolek [52]3 years ago
6 0

wait so which one is it?

can some one please kindly tell me which is the right answer?

Anni [7]3 years ago
3 0
H is equal to and greater than 48 the last one because you must be equal to 48 inches or greater than 48 inches so = would be equal to and < would mean that you must be greater than
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X >7
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I need some help on the math problem as soon as possible
givi [52]

Answer:

A(1,6)

Step-by-step explanation:

SOOO, we're gonna graph it okay?

I graphed it, and A was the only one on the line.

I hope this helps

7 0
3 years ago
The following are the ages (years) of 5 people in a room: 14, 16, 16, 13, 23 A person enters the room. The mean age of the 6 peo
myrzilka [38]

Answer:

44.

Step-by-step explanation:

Since the mean of 14, 16, 16, 13, 23, x is 21, then we have 14 + 16 + 16 + 13 + 23 + x is equal to 126. Summing we have 14 + 16 + 16 + 13 + 23 = 82. 82 + x is equal to 126. X = 44.

4 0
2 years ago
Soap films and bubbles are colorful because the interference conditions depend on the angle of illumination (which we aren't cov
mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

2t=(m+\frac{1}{2} ).\frac{\lambda}{n}

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm

Hence the thickness difference should be

t_1-t_2=131.58-75.19=56.39 \ nm

7 0
2 years ago
Triangle QRS has been rotated 90° to create triangle TVU. Using the image below, prove that lines RS and VU have the opposite an
Zanzabum

Answer:

If the line RS has been rotated 90 degrees, then VU will be perpendicular to RS and the two slopes must be opposite and reciprocal, i.e. product of the two slopes will equal -1.

As a verification, we find the locations of V and U from rotations of R & S.

(actually, the triangle had been rotated -90&deg;, 90 &deg; clockwise)

Step-by-step explanation:

Slope RS, m1:

Slope VU, m2

Hence m1*m2=1*-1=-1, meaning that m1 and m2 are opposite (in sign) and are reciprocal to each other, as expected

6 0
3 years ago
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