Answer:
The area of triangles are 16 cm^2 and 49 cm^2
Step-by-step explanation:
we know that
If two figures are similar, the ratio of its perimeters is equal to the scale factor and the ratio of its areas is equal to the scale factor squared
Let
z ----> the scale factor
x ----> the area of the smaller triangle in square centimeters
y ----> the area of the larger triangle in square centimeters
we know that
so
-----> equation A
----> equation B
solve the system by substitution
substitute equation A in equation B
solve for y
Find the value of x
therefore
The area of triangles are 16 cm^2 and 49 cm^2
X= 55° because their is a 90° angle so if you subtract the 35° angle you get 55 as the rest of the 90° angle
The y-intercept of a function is that function's value when x=0.
.. f(0) = 4
.. g(0) = 0
These y-intercepts differ by 4.
If you want them to differ by zero (be the same), you have to move them to the same point.
Selection 1. Moves f(0) down 3 (to 1) and g(0) up 1 (to 1), so moves the y-intercepts to the same point. This is what you want, so this is an appropriate answer selection.
Selection 2. Moves f(0) up 4 (to 8) and g(0) down 1 (to -1). 8 and -1 are not the same point, so this is not the answer.
Selection 3. Moves f(0) down 2 (to 2) and g(0) up 1 (to 1). 2 and 1 are not the same point, so this is not the answer.
Selection 4. Moves f(0) up 3 (to 7) and g(0) down 6 (to -6). 7 and -6 are not the same point, so this is not the answer.
___
The appropriate choice is the 1st one.
f(x) - 3 and
g(x) + 1
The intersection line of two planes is the cross product of the normal vectors of the two planes.
p1: z=4x-y-13 => 4x-y-z=13
p2: z=6x+5y-13 => 6x+5y-z=13
The corresponding normal vectors are:
n1=<4,-1,-1>
n2=<6,5,-1>
The direction vector of the intersection line is the cross product of the two normals,
vl=
i j k
4 -1 -1
6 5 -1
=<1+5, -6+4, 20+6>
=<6,-2,26>
We simplify the vector by reducing its length by half, i.e.
vl=<3,-1,13>
To find the equation of the line, we need to find a point on the intersection line.
Equate z: 4x-y-13=6x+5y-13 => 2x+6y=0 => x+3y=0.
If x=0, then y=0, z=-13 => line passes through (0,0,-13)
Proceed to find the equation of the line:
L: (0,0,-13)+t(3,-1,13)
Convert to symmetric form:
=>
