![\boxed{\overline{VT}=25\ units}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Coverline%7BVT%7D%3D25%5C%20units%7D)
<h2>
Explanation:</h2>
In order to find:
![\overline{VT}](https://tex.z-dn.net/?f=%5Coverline%7BVT%7D)
We'll use the Distance Formula:
![d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%29%5E2%2B%28y_%7B2%7D-y_%7B1%7D%29%5E2%7D)
In this case, let's call:
![V(x_{1},y_{1}) \rightarrow V(-1,-10) \\ \\ T(x_{2},y_{2}) \rightarrow T(6,14)](https://tex.z-dn.net/?f=V%28x_%7B1%7D%2Cy_%7B1%7D%29%20%5Crightarrow%20V%28-1%2C-10%29%20%5C%5C%20%5C%5C%20T%28x_%7B2%7D%2Cy_%7B2%7D%29%20%5Crightarrow%20T%286%2C14%29)
Then, applying distance formula between these two points:
![\overline{VT}=\sqrt{(6-(-1))^2+(14-(-10))^2} \\ \\ \overline{VT}=\sqrt{(6+1)^2+(14+10)^2} \\ \\ \overline{VT}=\sqrt{7^2+24^2} \\ \\ \overline{VT}=\sqrt{49+576} \\ \\ \overline{VT}=\sqrt{625} \\ \\ \boxed{\overline{VT}=25 \ units}](https://tex.z-dn.net/?f=%5Coverline%7BVT%7D%3D%5Csqrt%7B%286-%28-1%29%29%5E2%2B%2814-%28-10%29%29%5E2%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BVT%7D%3D%5Csqrt%7B%286%2B1%29%5E2%2B%2814%2B10%29%5E2%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BVT%7D%3D%5Csqrt%7B7%5E2%2B24%5E2%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BVT%7D%3D%5Csqrt%7B49%2B576%7D%20%5C%5C%20%5C%5C%20%5Coverline%7BVT%7D%3D%5Csqrt%7B625%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7B%5Coverline%7BVT%7D%3D25%20%5C%20units%7D)
<h2>Learn more:</h2>
Distance Formula: brainly.com/question/10134840
#LearnWithBrainly
Step-by-step explanation:
1/3 2x-24=16
2x-24=16×3
2x-24=48
2x=48+24
2x =72
x=72/2
x=36
Answer:
224.91 rounded to the nearest whole is 225 because it is closer to 225 than 224.
Step-by-step explanation:
Hope this helps!!
Pls give brainliest!!
Answer:
23/25 when reduced to the simplest form.
Step-by-step explanation:
write down the number as a fraction of one
0.92 = 0.92/1
so we multiply both numerator and denominator by 100
Then 0.92/1 = ( 0.92x100) / ( 1x100)
= 92/100
Now reduce the numerator and denominator by GCD between in this case
GCD ( 92,100) = 4
so, 92/ 4
100/ 4
=23/25
Answer:
Population of fish in 2010 = 1,646 fish (Approx.)
Step-by-step explanation:
Given:
Population of fish in 2002 = 2100 fish
Decrease rate = 3% per year
Find:
Population of fish in 2010
Computation:
Number of year = 2010 - 2002
Number of year = 8 year
A = P[1+r]ⁿ
Population of fish in 2010 = Population of fish in 2002[1+(-3%)]ⁿ
Population of fish in 2010 = 2100[1-0.03]⁸
Population of fish in 2010 = 2100[0.97]⁸
Population of fish in 2010 = 2100[0.7837]
Population of fish in 2010 = 1645.77
Population of fish in 2010 = 1,646 fish (Approx.)