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KiRa [710]
1 year ago
12

If the distance traveled is (4s3 - 6s² + 4s + 14) miles and the rate is (s + 1) mph, write an expression, in hours, for the time

traveled.​
Mathematics
1 answer:
BartSMP [9]1 year ago
3 0

The equation for time is URM:

time =  \frac{distance}{speed \: (rate)}

t(s) =  \frac{4s {}^{3}  - 6s {}^{2}  + 4s + 14}{s + 1}  =  \frac{2(s + 1)(2s {}^{2} - 5s + 7) }{s + 1}  \\  t(s) = 2(2s {}^{2}  - 5s + 7) \\ t(s) = 4s {}^{2}  - 10s + 14

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5c+ 3 < 28 or -4c-2< 14 for c.
sweet-ann [11.9K]

Answer:

c<5, and  c>-4.

<u>Steps used for answers:</u>

5c+3<28 =

1. Subtract 3 from both sides.

5c<28-3

2. Simplify  28-3 to  25.

5c<25

3. Divide both sides by 5.

c<25/5

4. Simplify  25/5 to 5.

c<5

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

-4c-2<14 =

1. Add 2 to both sides.

-4c<14+2

2. Simplify  14+2 to 16.

-4c<16

3. Divide both sides by -4.

c>-16/4

4. Simplify 16/4 to 4.

c>−4

<u>Done by NeighborhoodDealer</u>

3 0
3 years ago
Read 2 more answers
The graph of f(x) = x^2 has been shifted into the form f(x) = (x − h)^2 + k
stepan [7]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\\\&#10;% left side templates&#10;\begin{array}{llll}&#10;f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}&#10;\\ \quad \\&#10;f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}&#10;\end{array}\\\\&#10;--------------------\\\\

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\&#10;\left. \qquad   \right.  \textit{reflection over the x-axis}&#10;\\\\&#10;\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\&#10;\left. \qquad   \right.  \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\bullet \textit{ vertical shift by }{{  D}}\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind, let's see, it went to the right 2 units, and then up 3 units.

that simply means, C = -2, D = 3.
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