Sum of 2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=

for x²-4x+16=0
x=

x=

x=

x=

x=

x=

the roots are
x=-4 and 2+2i√3 and 2-2i√3
Area of the figure = 30.28 m²
Solution:
The given image is splitted into two shapes.
One is rectangle and the other is semi-circle.
Length of the rectangle = 6 m
Width of the rectangle = 4 m
Area of the rectangle = length × width
= 6 m × 4 m
= 24 m²
Area of the rectangle = 24 m²
Diameter of the semi-circle = 4 m
Radius of the semi-circle = 4 m ÷ 2 = 2 m
Area of the semi-circle = 


Area of the semi-circle = 6.28 m²
Area of the figure = Area of the rectangle + Area of the semi-circle
= 24 m² + 6.28 m²
= 30.28 m²
Area of the figure = 30.28 m²
Answer:
I'm pretty sure its D
Step-by-step explanation: Brainliest???
Answer:
<h2>x=29</h2>
Step-by-step explanation:
<h3>to understand the solving steps</h3><h3>you need to know about:</h3>
- right angle
- equation
- PEMDAS
<h3>to solve:</h3><h3>x</h3><h3>let's solve:</h3>
<u>a </u><u>right</u><u> angle</u><u> </u><u>contains </u><u>9</u><u>0</u><u>°</u> (always)
according to the question
x+3+x-1+x+1=90
3x3=90
x=87
x=29
<h2>solved</h2>
That's the answer in my opinion.good luck!