In the diagram, AB = 10 and AC = 2√10. What is the perimeter of △ABC?

Mathematics

1 answer:

Zigmanuir [339]3 years ago

4 0

D. 20 + 2âš10 units To solve this, you simply need to calculate the length of each side of the triangle with the vertexes of A(3,4), B(-5,-2), and C(5,-2). The length of each side is simply calculated using the pythagoras theorem. Note that it doesn't matter what order you do the subtraction. The absolute value will be the same and if it happens to be negative, not a problem since it will become positive once you square the values. So the length of side AB is sqrt((3-(-5))^2 + (4-(-2))^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10. The length of side BC is sqrt((-5 - 5)^2 + (-2 - (-2))^2) = sqrt(-10^2 + 0^2) = sqrt(100+0) = sqrt(100) = 10. And finally, the length of side AC is sqrt((3-5)^2 + (4-(-2))^2) = sqrt(-2^2 + 6^2) = sqrt(4+36) = sqrt(40) = 2 * sqrt(10) Finally, add all the lengths together. 10 + 10 + 2âš10 = 20 + 2âš10

You might be interested in

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

HELPPP ASAPPPPPP PLEASEEEE !!!!!

Verizon [17]

The correct answer is B

8 0

3 years ago

What is the length of the radius of a circle with a center at 2 + 3i and a point on the circle at 7 + 2i?

dem82 [27]

You can calculate the difference/vector of both points:
$7+2i-(2+3i)=5-i$
then calculate the magnitude/length like pythagoras
$radius=\sqrt{5^2+(-i*-i)}\\=\sqrt{5^2+i^2}\\=\sqrt{5^2-1}\\=\sqrt{24}\\=\sqrt{4*6}\\=2*\sqrt{6}$