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Gre4nikov [31]
3 years ago
13

Jamaal bounces on a trampoline. His height, as a function of time, is modeled by y = -6x2 + 20x +4

Mathematics
2 answers:
kykrilka [37]3 years ago
5 0

Answer: The function is no linear

Step-by-step explanation:

Butoxors [25]3 years ago
5 0

Answer: the func is nonlinear

Step-by-step explanation:

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PLZ HELP!!!<br><br> Plz answer fast if possible.
wlad13 [49]

Answer:

For the first one:

yes

constant rate of 4 inches of perimeter/per inch of side length

For the second:

no

not a constant rate

6 0
3 years ago
Which of the follow is a factor of 3x^3+18x^2+27x
blagie [28]
3x(x+3)^2 is your answer
start 3x^3+18x^2+27x
Pull out common factor of 3x
3x(x^2+6x+9)
Factor into 2 (x+3)s
then <span>combine
</span>3x(x+3)^2
Your factors are 3x, (x+3) and (x+3)

Hope this helps. If it did, make it brainliest
3 0
3 years ago
Read 2 more answers
Question five for geometry quiz
Helga [31]

The two tangents are parallel to each other because the two lines perpendicular to the same line are parallel to each other.

Lk + JK = JL




The answer is Parallel to each other
3 0
3 years ago
A model for the density δ of the earth’s atmosphere near its surface isδ=619.09−0.000097pwhere p (the distance from the center o
natta225 [31]

Answer: 3.751*10^18kg

Step-by-step explanation:

δ =619.09−0.000097p....equa1 where p (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter.

Calculating the density of air at 5km above earth surface

P = 5000m + 6370000m = 6.375*10^6m

δ = 619.09 -(.000097* 6.375*10^6)

δ = 0.715kg/m^3 = density

Since Mass = density*volume...equ2

To calculate volume of air around the spherical earth at height 5km

V = (4/3 pai R^3) - (4/3pai r^3) ...equation 3 where R =6.375*10^6m, r = 6.37*10^6

Substituting R and r in equation 2 to solve for volume of air

V = 1.085*10^21 - 1.08*10^21

V = 5.25*10^18m^3

Substituting δ and V into equation 2 to solve for mass of air

M = 0.715 * (5.25*10^18)

M = 3.751*10^18kg

8 0
3 years ago
Answer fast pleaseeee :)
pochemuha

<u>Answer:</u>

The correct answer option is 'both Adler and Erika solved for <em>k</em> correctly because either the addition property of equality or the subtraction property of equality can be used to solve for <em>k</em><em>'.</em>

<u>Step-by-step explanation:</u>

Both Adler and Erika have solved for <em>k</em> correctly since both the properties, addition property of equality and subtraction property of equality, can be to solve for <em>k.</em>

The fraction \frac{1}{2} has to be subtracted from both the sides so even if it is added, the negative sign still would make it subtracted from both the sides.

8 0
3 years ago
Read 2 more answers
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