Answer:
ASA
Step-by-step explanation:
You can show the angles at either end of segment BC in triangles MBC and LCB are congruent, so you have two angles and the segment between. The appropriate theorem in such a case is ASA.
-3...idk what your asking...but on a number line, 3 places to the left of zero is negative three
9514 1404 393
Answer:
34.5 square meters
Step-by-step explanation:
We assume you want to find the area of the shaded region. (The actual question is not visible here.)
The area of the triangle (including the rectangle) is given by the formula ...
A = 1/2bh
The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...
A = (1/2)(11 m)(9 m) = 49.5 m^2
The rectangle area is the product of its length and width:
A = LW
The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...
A = (5 m)(3 m) = 15 m^2
The shaded area is the difference between the triangle area and the rectangle area:
shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2
The shaded region has an area of 34.5 square meters.
Answer: x = -0.377
Step-by-step explanation:
We have the equation:
4^(5*x) = 3^(x - 2)
Now we can use the fact that:
Ln(A^x) = x*Ln(A)
Then we can apply Ln(.) to both sides of the equation to get:
Ln(4^(5*x)) = Ln(3^(x - 2))
(5*x)*Ln(4) = (x - 2)*Ln(3)
(5*x)*Ln(4) - x*Ln(3) = -2*Ln(3)
x*(5*Ln(4) - Ln(3)) = -2*Ln(3)
x = -2*Ln(3)/(5*Ln(4) - Ln(3)) = -0.377
Answer:
A) 25
B) 4
C) 0.25
Thank you and please rate me as brainliest as it will help me to level up
