Solving for the missing term and the missing coefficient (6a − )5a = ( ) a^2 − 35a
Let the missing term be X
Let the missing coefficient be Y
Therefore, (6a – X)5a = Y(a^2) – 35a
6a x 5a – X.5a = Y.a^2 – 35a
30a^2 – X.5a = Y.a^2 – 35a
Equating co-efficients,
30a^2 = Y.a^2; X.5a = 35a
30 = Y; 5X = 35
Y = 30; X = 7
Therefore, (6a-7)5a = 30 a^2 – 35a

3 0

3 years ago

Explain the difference between a square pyramid and a square prism.

Maslowich

The square pyramid looks exactly how it sounds. The base of the shape is a square and has 4 isosceles triangles that meet at a point. The square prism is like a 3D rectangle. It has 4 rectangles and 2 squares to close the shape.
Hope this helped.

3 0

3 years ago

Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$