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Rina8888 [55]
3 years ago
5

Which of the following best describes the solution to the equation below?

Mathematics
1 answer:
kozerog [31]3 years ago
4 0
D)One solution. b=5
Distribute 2 times b plus 2 times 3
2b+6
2b+6+2b=26
Combine like terms 4b+6=26
Subtract 6 from both sides
4b=20
Divide both sides by 4
b=5

Check by substituting 5 for b
2 (5+3)+2(5)=26
2(8) + 2(5)=26
16+10=26
26=26
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Answer:

A. 40/100

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On two tests so far this year, a student received a 78 and a 93. The student wants an average between 80 and 90 inclusive. What
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<span>240 <= 171 + x <= 270
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By car, John traveled from city A to city B in 3 hours. At a rate that was 20 mph
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Not good at physics problems, need help!
xxMikexx [17]

Next time, please post in the physics section. It is not exactly a math problem because it requires knowledge of kinematics equations, assumptions and constants.

Given:

A rocket fired (assumed vertically upwards) at an initial velocity of +123 ft./s (positive upwards) against acceleration due to gravity of (assumed -32.2 ft/s^2) at an initial height of +4 feet above the ground. Need to find the time it takes before rocket reaches ground (assuming air resistance negligible) due to free fall.

Kinematics equations:

Using given numerical information

u=inivitial velocity=+123 ft/s

g=acceleration due to gravity= -32.2 ft/s^2

s0=initial location, height=+4 ft

s1=final local, height = 0 ft (ground)

We make use of the kinematics equation

s1=s0+u*t+at^2.............................(1)

where a=acceleration = g = -32.2 ft/s^2

Subsitute values in (1)

0=4+123*t-32.2t^2

and solve for t using the quadratic formula (knowing that t>0)

t=[-123 &pm; sqrt(123^2-4*(-32.2)*4)]/(2*-32.2) =>

t = -0.03225 sec. or t=3.852 sec.

Reject negative root, so the rocket will hit the ground 3.852 seconds since airborne.

Note: since #61 and #62 are pale in colour, it will be assumed that you can handle those two, or they will be the subject of other questions. Thank you.

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3 years ago
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