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Zepler [3.9K]
3 years ago
13

A map is drawn using a scale of 80 mi to 1 cm. On the map , the two cities are 7.5 cm apart. What is the actual distance between

the two cities ??
Mathematics
1 answer:
meriva3 years ago
6 0
We know that
scale =measure in the map/measure in the actual
measure in the actual=measure in the map/scale
scale=1 cm/ 80 mi
for
measure in the map=7.5 cm
measure in the actual=?
measure in the actual=7.5 /(1/80)-----> 7.5*80-----> 600 mi

the answer is
<span>the actual distance between the two cities is 600 miles</span>
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Tryna figure this out
kaheart [24]

Answer:

9pi approximately 28

Step-by-step explanation:

we have the formula - pi r^2

which we know the diameter is 6 m and the surface is a semi circle

the radius is 3 so then we have pi 3^2

which becomes 9pi which is about 28.2

rounding to 28

hope this helps!!

7 0
3 years ago
Julie is constructing a scale model of her room.The rectangular room is 10 1/4 inches by 8 inches.If 1 inch represents 2 feet of
zlopas [31]
The way to determine this is to know that 1 foot is 12 inches (so 2 is 24) 
Now the ration that would determine the scale factor of the room is 1:24 (1 inch for every 24 inches)
So the scale factor is 1:24
Now to determine the area we multiply the numbers we have by 2 and change the inches to feet (I hope that makes sense to you, it does to me, I'll show you)
10.25 * 2 = 20.5 ft.
8 * 2 = 16 ft.
now we know the dimensions of the room so we need to find the area.
A=B*H
20.5 * 16 = 328
so the area of the room is 328 ft.²
6 0
3 years ago
Read 2 more answers
How many ml of 20 percent acid should be added to pure acid to make 70 ml of 30 percent acid?
Novay_Z [31]

Answer:

61.25%

Step-by-step explanation:

How many ml of 20 percent acid should be added to pure acid

to make 70 ml of 30 percent acid?

:

let x = amt of 20% mixture

then

(70-x) = amt of pure acid

:

0.20x + (70-x) =0 .30(70)

0.20x + 70 - x = 21

= 21 - 70

-0.8x =-49

x = 61.25%

x = +61.25 ml of 20% stuff

8 0
3 years ago
P = a^b + b^a. Given a,b,p are prime numbers. find the value of p
VARVARA [1.3K]
2³   +  3² = 2*2*2+3*3= 8+9=17
a=2,
b=3,
P= 17,
2,3,17 are all prime numbers

4 0
3 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height
Crank

Answer:

a) h = 0.1: \bar v = -11\,\frac{ft}{s}, h = 0.01: \bar v = -10.1\,\frac{ft}{s}, h = 0.001: \bar v = -10\,\frac{ft}{s}, b) The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

Step-by-step explanation:

a) We know that y = 30\cdot t -10\cdot t^{2} describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (\bar v), measured in feet per second, can be done by means of the following definition:

\bar v = \frac{y(2+h)-y(2)}{h}

Where:

y(2) - Position of the ball evaluated at t = 2\,s, measured in feet.

y(2+h) - Position of the ball evaluated at t =(2+h)\,s, measured in feet.

h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

h = 0.1\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}

y(2.1) = 18.9\,ft

\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}

\bar v = -11\,\frac{ft}{s}

h = 0.01\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}

y(2.01) = 19.899\,ft

\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}

\bar v = -10.1\,\frac{ft}{s}

h = 0.001\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}

y(2.001) = 19.99\,ft

\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}

\bar v = -10\,\frac{ft}{s}

b) The instantaneous velocity when t = 2\,s can be obtained by using the following limit:

v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}

v(t) =  \lim_{h \to 0} 30-20\cdot t-10\cdot h

v(t) = 30\cdot  \lim_{h \to 0} 1 - 20\cdot t \cdot  \lim_{h \to 0} 1 - 10\cdot  \lim_{h \to 0} h

v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at t = 2\,s:

v(2) = 30-20\cdot (2)

v(2) = -10\,\frac{ft}{s}

The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

8 0
3 years ago
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