the equation of a line in point-slope form is
y - b = m(x - a)
where m is the slope and (a, b) a point on the line
to calculate m use the gradient formula
m = (y₂ - y₁ ) / (x₂ - x₁ )
with (x₁, y₁ ) = (- 4, - 1) and (x₂, y₂) = (1 1/2, 2 )
m = 
= 
= 
using (a, b) = (- 4, - 1), then
y + 1 = (x + 4)
Answer:
∠x = 90°
∠y = 58°
∠z = 32°
Step-by-step explanation:
he dimensions of the angles given are;
∠B = 32°
Whereby ΔABC is a right-angled triangle, and the square fits at angle A, we have;
∠A = 90°
∠B + ∠C = 90° which gives
32° + ∠C = 90°
∠C = 58°
∠x + Interior angle of the square = 180° (Sum of angles on a straight line)
∠x + 90° = 180°
∠x = 90°
∠x + ∠y + 32° = 180° (Sum of angles in a triangle)
90° + ∠y + 32° = 180°
∠y = 180 - 90° - 32° = 58°
∠y + ∠z + Interior angle of the square = 180° (Sum of angles on a straight line)
58° + ∠z +90° = 180°
∴ ∠z = 32°
∠x = 90°
∠y = 58°
∠z = 32°
The quadratic formula is above.
You want each equation in standard form: ax^2 + bx + c = 0
I begin each problem by defining variables.
For instance 3. Is in standard form. X^2 -2x - 3 = 0. a = 1, b = -2, c = -3
Now use quadratic formula: x = [- b + or - sqrt(b^2 - 4ac)]\2a
. The series is divergent. To see this, first observe that the series ∑ 1/kn for n = 1 to ∞ is divergent for any integer k ≥ 2.
Now, if we pick a large integer for k, say k > 100, then for nearly all integers n it will be true that 1 > cos(n) > 1/k. Therefore, since ∑ 1/kn is divergent, ∑ cos(n)/n must also be divergent The *summation* is divergent, but the individual terms converge to the number 0.<span>by comparison test since cosn/n <= 1/n is convergent
and 1/n is divergent by harmonic series
so the series is conditionally converget </span>
Use the tens.Try 30x80. 76 rounds up to 80 and 33 rounds down to 30.
