Answer: 3∛2
<u>Step-by-step explanation:</u>
First, factor 128 .... you need three common factors to bring one on the outside of the radical.
![\dfrac{3}{4}\sqrt[3]{128} \\\\\\=\dfrac{3}{4}\sqrt[3]{4\cdot 4\cdot 4\cdot 2} \\\\\\=\dfrac{3}{4}\cdot 4\sqrt[3]{2} \\\\\\=\large\boxed{3\sqrt[3]{2} }](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7D%7B4%7D%5Csqrt%5B3%5D%7B128%7D%20%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B3%7D%7B4%7D%5Csqrt%5B3%5D%7B4%5Ccdot%204%5Ccdot%204%5Ccdot%202%7D%20%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B3%7D%7B4%7D%5Ccdot%204%5Csqrt%5B3%5D%7B2%7D%20%5C%5C%5C%5C%5C%5C%3D%5Clarge%5Cboxed%7B3%5Csqrt%5B3%5D%7B2%7D%20%7D)
Part A;
There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.
An example of such system of equation is
x < 0
y > 0
The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.
Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.
Substituting D(-4, 2) into the system
we have:
-4 < 0
2 > 0
as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.
Also, substituting E(-1, 5) into the system we have:
-1 < 0
5 > 0
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.
To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.
For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true
For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false
For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false
For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true
For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true
For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true
Therefore, the farms in which chicken can be raised are the farms at point A, D, E and F.
You can use the Pythagorean Theorem to find the length of the third side AB (Identified as "x" in the figure attached in the problem), which says that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:
a² = b²+c²
As we can see the figure, the triangle does not have an angle of 90°, but it can be divided into two equal parts, leaving two triangles with a right angle. We already have the values of the hypotenuse and a leg in triangle "A" , so we can find the value of the other leg:
b = √(a²-c²) b = √(10²-4²) b = 9.16
With these values, we can find the hypotenuse in the triangle "B": x = √b²+c² x = √(9.16)²+(4)² x = 10
Answer: The boxer lose 1.8 kg in the final week to get able to compete as a flyweight.
Step-by-step explanation:
Since we have given that
Quantity of weight a boxer needs to lose in a month is given by
In 3 months, he lowers his weight from 55.5 kg to 53.8 kg.
So, Quantity of weight he lose in three months is given by
Number of kilograms the boxer lose in the final week to be able to compete as a flyweight is given by
Hence, the boxer lose 1.8 kg in the final week to get able to compete as a flyweight.
