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3 years ago

The table shows the distance in feet, y, that a rocket traveled in x seconds.

Mathematics

2 answers:

Murrr4er [49]3 years ago

7 0

Answer:

The correct option is D. y = 25 $x^{2}$ .

Step-by-step explanation:

i) From the table and our own observation and a trial and error approach we

can clearly see that the equation that matches y in feet ( the distance

traveled by the rocket) and x in seconds ( the time elapsed) is given by the

equation y = 25 $x^{2}$

Therefore the correct option is D. y = 25 $x^{2}$ .

brilliants [131]3 years ago

4 0

Answer:

The correct option is D. y = 25.

Step-by-step explanation:

i) From the table and our own observation and a trial and error approach we

can clearly see that the equation that matches y in feet ( the distance

traveled by the rocket) and x in seconds ( the time elapsed) is given by the

equation y = 25

Therefore the correct option is D. y = 25.

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Ede4ka [16]

in the number line, the end points are DG, and the point in between is O

DG = 88

DO = 5x + 12

OG = 2x

Set the equation. The two parts (DO & OG) are equal to the whole (DG)

2x + 5x + 12 = 88

Simplify. Combine like terms

(2x + 5x) + 12 = 88

7x + 12 = 88

Isolate the x. Remember to do the opposite of PEMDAS. Subtract 12 from both sides

7x + 12 (-12) = 88 (-12)

7x = 76

Isolate the x. Divide 7 from both sides:

7x/7 = 76/7

x = 76/7

----------------------------------------------------------------------------------------------------------------------------

Find DO. Plug in "76/7" for x:

DO = 5x + 12

DO = 5(76/7) + 12

Simplify. Remember to follow PEMDAS. Multiply 76 with 5

DO = 380/7 + 12

Next, divide 380 with 7

DO = 54.29 (rounded)

Finally, add

DO = 54.29 + 12

DO = 66.29

66.29 is your answer

hope this helps

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3 years ago

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zysi [14]

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pogonyaev

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Julian is using a biking app that compares his position to a simulated biker traveling Julian's target speed. When Julian is beh

zepelin [54]

Answer:

$\large \boxed{\text{11 km/h}}$

Step-by-step explanation:

1. The situation after 15 min:

(a) Distance travelled by simulated biker:

15 min =¼ h

$\text{Distance} = \dfrac{1}{4}\text{ h} \times \dfrac{\text{20 km}}{\text{1 h}} = \text{5 km}$

(b) Distance travelled by Julian

Julian is 2¼ km behind the biker. The distance he has travelled is (5 - 2¼) km

5 - 2¼ = 5 - ⁹/₄ = ²⁰/₄ - ⁹/₄ = ¹¹/₄ = 2¾

Julian has travelled 2¾ km in ¼ h.

2. Julian's average speed

$\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{2\frac{3}{4}\text{ km}}{\frac{1}{4}\text{ h}} =\dfrac{11}{4}\text{ km}\times \dfrac{4}{\text{1 h}} = \textbf{11 km/h}\\\\\text{Julian's average speed is $\large \boxed{\textbf{11 km/h}}$}$

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3 years ago

Read 2 more answers