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-Dominant- [34]
2 years ago
15

the scale of a map says that 4 cm represents5 km.What distance on the map in centimeters represents an actual distance of 4 kilo

meters
Mathematics
1 answer:
vladimir1956 [14]2 years ago
5 0

Answer:

3.2 cm

Step-by-step explanation:

If we know that our ratio is 4 cm: 5 km. We can find the unit rate of the scale. Which would be 0.8 cm: 1 km. After we find the unit ratio we can then multiply both sides by 4. (4 * 0.8) : (4*1) After simplifying we get the ratio of 3.2cm: 4 km.

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What is the surface area of the cylinder in terms of pi? The diagram is not drawn to scale.
bearhunter [10]
Surface Area = 2πr² + 2πrh 

Surface Area = 2π(7)² + 2π(7)(15) = 308π in²²

Answer: 308π in²
3 0
3 years ago
A train travels 210.6 miles in 1.3 hours what is the trains average speed?
grin007 [14]
Average speed = total distance/ total time
210.6/1.3= 162
6 0
3 years ago
Read 2 more answers
A) -10 &lt; y &lt; 10 <br><br> B) -2 &lt; y &lt; 7 <br><br> C) { -2, 1, 4, 7 } <br><br> D) { -2, 7 }
nekit [7.7K]
The answer to this should be b

7 0
3 years ago
..........................
QveST [7]

Answer:

x = 0.954

Step-by-step explanation:

We apply pythagorean to find x

(4-x)² = 1.2² + 2.8²

(4-x)² = 9.28

✓(4-x)² = ✓(9.28)²

4 - x = 3.046

- x = 3.046 - 4

- x = -0.954

x = 0.954

(Please heart and rate if you find it helpful, it's a motivation for me to help more people)

7 0
2 years ago
A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula
Ilia_Sergeevich [38]

Answer: A. A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}

              B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

A=\pi.r^{2}

So to find the rate of the area:

\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}

\frac{dA}{dr} =2.\pi.r

Using r(t)=3-\frac{363}{(t+11)^{2}}

\frac{dr}{dt}=\frac{726}{(t+11)^{3}}

Then

\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]

\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}

Multipying and simplifying:

\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}

The rate at which the area is increasing is given by expression A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}.

B. At t = 5, rate is:

A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}

A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}

A'(5)=\frac{2408693760\pi}{4294967296}

A'(5)=1.76268

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0
2 years ago
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