Answer:
No solution
Step-by-step explanation:
When we take the 5/2 x to the other side, 4x and 4x cancels out. We get 7=0 which is incorrect. Therefore, x has no solution.
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B(t)=50×e^2.5t
e^2.5t = 20
t= ln(20) / 2.5
set B(t)=1000 and solve for t via 1000=50e2.5t
20=e2.5t
ln(20)=2.5t
t=ln(20) / 2.5
Answer:
0.25
Step-by-step explanation:
Answer:
80 miles
Step-by-step explanation:
The map scale compares the distance between two points on the map with the corresponding distance on the ground.
That is,
![\boxed{\mathfrak{map\:scale=\frac{distance\:on\:map}{distance\:on\:ground} }}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cmathfrak%7Bmap%5C%3Ascale%3D%5Cfrac%7Bdistance%5C%3Aon%5C%3Amap%7D%7Bdistance%5C%3Aon%5C%3Aground%7D%20%7D%7D)
The formula makes it pretty easy to solve!
- map scale is <em>1 inch : 20 miles</em>
- distance on map is <em>4 inches</em>
- Let's say, the corresponding distance on the ground is <em>x miles.</em>
<em />
Substituting these values in the above mentioned formula:
<em>cross multiplying:</em>
![\implies\mathsf{x=4 \times 20}](https://tex.z-dn.net/?f=%5Cimplies%5Cmathsf%7Bx%3D4%20%5Ctimes%2020%7D)
![\implies \mathsf{x=80}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cmathsf%7Bx%3D80%7D)
<u>That makes the actual distance 80 miles!</u>
<u> </u>
<h3>ALTER</h3>
You can also use simple unitary method here:
1 inch on the map = 20 miles on the ground
∴ 4 inches on the map = 4 × 20 miles on the ground
<u>= 80 miles on the ground</u>
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