CPhill

Slope = change in y / change in x

So from the first point to the second, y changes by -3

And x changes by -3

So -3 / -3   = 1

And that's the slope

So we have

Area =(1/2)(Base)(Height)

A = (1/2)(8)(A - 6)

A = 4(A - 6)

A = 4a - 24    subtract 4a from both sides

-3A = -24      divide both sides by -3

A  = 8 sq. units

(12+34i) and (12−34i) divided by (2−3i)

(12 + 34i)(12 - 34i) =

144 - 1156i^2 =

144 - 1156(-1) =

144 + 1156 =

1300

And

1300 / (2-3i)    multiply top and bottom by the conjugate of 2 - 3i

1300(2 + 3i) / [ (2 - 3i)(2 + 3i) =

1300(2 + 3i) / ( 4 - 9i^2)

1300(2 + 3i) / 13 =

2600/13 + 3900i / 13 =

200 + 300i

Thanks, h7......!!!

99*54/12+96-(75-61)

Multiply 99*54 = 5346

5346/12 = 445.5

445.5 + 96 = 541.5

So

541.5 - (75 - 61)  =

541.5 - (14) =

527.5

We have SSA (the ambiguous case) here..

Using the Law of Sines, we have

sinA / a = Sin B / b

sinA / 7 = sin40 / 5

sinA = (7/5)sin40

sin^-1((7/5)sin 40) =A = about 64.15 degrees

So, angle C is 180 - 40 - 64.15 = 75.85 degrees

And side C is given by

a / sin A = c / sin C

5 / sin 40 = c / sin 75.85 degrees

c = 5 sin75.85 / sin 40  = 7.54

But....angle A could also be a second quadrant angle = 180 - 64.15 = 115.85

So the other possible angles are 40, 115.85 and 24.15

And side c in this triangle is given by

5 / sin 40 = c / sin 24.15

c = 5sin24.15/sin40 = 3.18

So the two triangle are (sides, angles)

a = 7, b = 5,  c = 7.54    A = 64.15, B = 40, C = 75.85   and

a = 7, b = 5, c = 3.18    A= 115.85, B = 40, C = 24.15

(64^4)^2 / (8^2)^7  simplify

64 ^8 / 8^14       note, 64 = 8^2

(8^2)^8 / 8^14  =

8^16 / 8^14 =

8^(16 - 14)  =

8^2 =

64

Is there supposed to be something to solve here???

100x^2 - 40x + 4 = 1    subtract 1 from both sides

100x^2 - 40x + 3  = 0   factor

(10x -3)(10x -1) = 0

So, setting each factor to 0, we have

10x - 3 = 0   →  10x = 3 →    x = 3/10

10x - 1 = 0   →  10x = 1 →    x = 1/10

"a" is correct

Quadrillion.....Quintillion......Sextillion......Septillion, etc.