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Anarel [89]
3 years ago
14

A car travels at constant speed for 7 days and covers 1500 miles. What is the speed in centimeters/minute? You may use the (appr

oximate) conversion factors on page 55 of the text (in particular that 1 mile is about 8/5 km), or you may relate everything to the conversion 1 inch = 2.54cm.
Mathematics
2 answers:
svetoff [14.1K]3 years ago
6 0

     (1500 mi / 7 da) · (1 da / 1440 min) · (1609.3 m / 1 mi) · (100 cm / 1 m)

  =  (1500 · 1 · 1609.3 · 100) / (7 · 1440 · 1 · 1)    cm/min

  =        23,948.6 cm/min
zvonat [6]3 years ago
3 0

Answer:

Speed = 23809.52 cm/min      

Step-by-step explanation:

Given : A car travels at constant speed for 7 days and covers 1500 miles.

To find : What is the speed in centimeters/minute

Solution : Formula to find speed is

\text{Speed}=\frac{\text{Distance}}{\text{Time}}

Distance traveled = 1500 miles and time taken = 7 days

\text{Speed}=\frac{\text{1500 miles}}{\text{7 days}}

Now, we have to convert miles into centimeter and days into minutes.

1 day = 24 hours

1 hour= 60 minute

24 hr= 24×60 minute

1 day =  24×60 minute

7 days = 7×24×60 minute = 10,080 minute

Now, 1 mile = 8/5 km = 1.6 km

1 km= 100000 cm

1.6 km = 1.6 × 100000 cm

1 mile = 1.6 × 100000 cm

1500 miles = 1500×1.6 × 100000 cm = 240,000,000 cm

Put back the values into speed

\text{Speed}=\frac{\text{1500 miles}}{\text{7 days}}

\text{Speed}=\frac{\text{240,000,000 cm}}{\text{10,080 minute}}

\text{Speed}=\text{23809.523 cm/minute}

Therefore, speed = 23809.52 cm/min

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An open topped box is constructed as follows: A 24 in. x 24 in. square bottom has four identical squares cut off at each corner.
Arte-miy333 [17]

Answer:

Step-by-step explanation:

Given

square bottom area is 24\times 24\ in.^2

Suppose x in. is cut from each corner to make a open box with maximum volume

New base area is (24-x)\times (24-x) in.^2

Volume of box

V=(24-x)^2\times x

Differentiate V w.r.t x to get maximum volume

\frac{\mathrm{d} V}{\mathrm{d} x}=2\cdot (24-x)(-1)+1\cdot (24-x)^2

Put \frac{\mathrm{d} V}{\mathrm{d} x}=0

\left ( 24-x\right )\left [ -2x+24-x\right ]=0

\left ( 24-x\right )\left [ 24-3x\right ]=0

x=24,8

but x=24 is not possible therefore x=8 will yield maximum volume

V=(24-8)^2\cdot 8=2048\ in.^2    

7 0
3 years ago
I am not sure how to solve this could anyone help???
faltersainse [42]

Answer:

about 137 degrees

Step-by-step explanation:

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6 0
3 years ago
Before the coffee shop opens the employees bake 40 cookies which is 25 less than the number of donuts they make does the equatio
Andre45 [30]

Answer:

x-25 = 40

Step-by-step explanation:

Let the number of cookies made = x

The employees bake 40 cookies which is 25 less than the number of donuts they make

Number of Cookies is less than Number of Donuts by 25

Therefore the equation that describes the situation is given as:

x-25 = 40 where x is the number of cookies made.

8 0
3 years ago
Read 2 more answers
Dy/dx = 2xy^2 and y(-1) = 2 find y(2)
Anarel [89]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2887301

—————

Solve the initial value problem:

   dy
———  =  2xy²,      y = 2,  when x = – 1.
   dx


Separate the variables in the equation above:

\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}


Integrate both sides:

\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}

\mathsf{\dfrac{1}{y}=-(x^2+C_1)}


Take the reciprocal of both sides, and then you have

\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}


In order to find the value of  C₁  , just plug in the equation above those known values for  x  and  y, then solve it for  C₁:

y = 2,  when  x = – 1. So,

\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}

\mathsf{C_1=-\,\dfrac{3}{2}}


Substitute that for  C₁  into (i), and you have

\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}


So  y(– 2)  is

\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0
3 years ago
9 times what number minus 7 equals negative seven
kirill115 [55]
9x-7=-7
9x=o
x=0
so the answer....is 0
8 0
3 years ago
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