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Natasha2012 [34]
3 years ago
5

Bonita is testing prototype model rocket engines. To be considered a successful launch, the rocket must reach a height of at lea

st 24 ft and must go a horizontal distance of no more than 10 ft. The vertical height must be at least three times as high as the horizontal distance. No rocket should go higher than 33 ft.
The graph shows the feasible region, where x represents the horizontal distance and y represents the vertical distance.

Which ordered pairs meet all the constraints for a successful launch and make sense in context of the situation?

Select each correct answer.




(0, 33)

(4.8, 30.5)

(9, 26)

(4, 36)

(2, 22)

Mathematics
1 answer:
lbvjy [14]3 years ago
5 0

Answer:

(0, 33)   : Yes

(4.8,  30.5)  : Yes


Step-by-step explanation:

height of at least 24 ft :  y >= 24

horizontal distance of no more than 10 ft. : 0 <= x <= 10

The vertical height must be at least three times as high as the horizontal distance. No rocket should go higher than 33 ft.:

y >=  3x and y <= 33

So Constraints:

24 <= y <= 33

y >= 3x

0 <= x <= 10

Now check to see if those coordinate points meet all the constraints

(0, 33)   : Yes

(4.8, 30.5)  : Yes

(9, 26)  : No (because 26 is less than (3 * 9) = 27)

(4, 36) : No (because 36 > 33,  must be 24 <= y <= 33)

(2, 22): No (because 22 <24,  must be 24 <= y <= 33)

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Answer:

60

Step-by-step explanation:

<u>Step 1:  Solve for x</u>

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<u>Step 2:  Find the measure of angle C</u>

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3 0
4 years ago
Find all the missing dimensions when a=9, b=13, c=64°
DerKrebs [107]

Answer:

Part 1) c=12.14\ units

Part 2) m\angle A=41.78^o

Part 3) m\angle B=74.22^o

Step-by-step explanation:

step 1

Find the measure of side c

Applying the law of cosines

c^2=a^2+b^2-2(a)(b)cos(C)

we have

a=9\ units\\b=13\ units\\C=64^o

substitute

c^2=9^2+13^2-2(9)(13)cos(64^o)

c^2=81+169-234cos(64^o)

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c^2=147.4212

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step 2

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Applying the law of sines

\frac{a}{sin(A)}=\frac{c}{sin(C)}

substitute the given values

\frac{9}{sin(A)}=\frac{12.14}{sin(64^o)}

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sin(A)=0.6663

m\angle A=sin^{-1}(0.6663)=41.78^o

step 3

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we know that

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so

m\angle A+m\angle B+m\angle C=180^o

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lukranit [14]

Answer:

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Answer:

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