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OleMash [197]
4 years ago
8

What is the answer for this question?

Mathematics
2 answers:
Tom [10]4 years ago
6 0

Answer:

Option C

Step-by-step explanation:

In the slope-intercept form of linear equation

y = m*x + b [tex]"m" indicates the slope (meaning how fast y varies with respect to x)"b" indicates the y-intercept of the line (at point (0,b))In this exercise, we are asked for the fastest variation of y with respect to x. In other words, the bigger absolute value of the slope, the steeper the graph. Notice that, taking absolute value, we cover both cases when the line is incresing (positive slope) and when the line is decreasing (negative slop)Let's do the calculations...[tex]y = 4x - 3 => |slope| = |4| = 4 y = (-3/4)x + 5 => |slope| = |-3/4| = 3/4 = 0.75y = (1/2)x + 2 => |slope| = |1/2| = 1/2 = 0.5 (least steep)y = 10x - 8 => |slope| = |10| = 10 Comparing the values, we conclude that y = (1/2)x + 2 has the least steep graph

const2013 [10]4 years ago
3 0

The least steep graph is y=1/2 x+2, or C, because the slope of the graph is the closest to 0, or being a flat horizontal line.

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explanation
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  !   Inches        !     Inches    !  Pounds   !       Pounds      !
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  !       0,9          ! 0,9*3 = 2,7 !        10     !                         !
  !       0,9          !                   !        10      !      3*10 = 30   !
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Answer:

(a) The population after 15 years is 2678.

(b)Therefore the population P(t) at any time t>0 is

P(t)= 45t+30 {t^{\frac32}}+260

Step-by-step explanation:

Given that,

The population grew at a rate of

P'(t)=45(1+\sqrt t)

Integrating both sides

\int P'(t) dt=\int 45(1+\sqrt t)dt

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\Rightarrow P(t)= 45t+45\  \frac{t^{\frac12+1}}{\frac12+1}+c              [ c is integration constant]

\Rightarrow P(t)= 45t+45\  \frac{t^{\frac32}}{\frac32}+c

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P(t)= 45t+30 {t^{\frac32}}+260

To find the population after 15 years, we need to plug t=15 in the above expression.

P(15)=( 45\times 15)+30( {15^{\frac32}})+260

         ≈2678

The population after 15 years is 2678.

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