Answer:
Below in bold.
Step-by-step explanation:
300 degrees - reference angle is |360 - 300 |= 60 degrees
225 = 225 - 180 = 45 degrees
480 = 480 - 360 = 120 so it is 180 - 120 = 60 degrees.
-210 = |-210 + 180| = 30 degrees.
Answer:
D) 0 = 2(x + 5)(x + 3)
Step-by-step explanation:
Which of the following quadratic equations has no solution?
We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.
A) 0 = −2(x − 5)2 + 3
0 = -2(x - 5) × 5
0 = (-2x + 10) × 5
0 = -10x + 50
10x = 50
x = 50/10
x = 5
Option A has a solution of 5
B) 0 = −2(x − 5)(x + 3)
Take each of the factors and equate them to zero
-2 = 0
= 0
x - 5 = 0
x = 5
x + 3 = 0
x = -3
Option B has a solution by one of its factors as a positive value of 5
C) 0 = 2(x − 5)2 + 3
0 = 2(x - 5) × 5
0 = (2x -10) × 5
0 = 10x -50
-10x = -50
x = -50/-10
x = 5
Option C has a solution of 5
D) 0 = 2(x + 5)(x + 3)
Take each of the factors and equate to zero
0 = 2
= 0
x + 5 = 0
x = -5
x + 3 = 0
x = -3
For option D, all the values of x are 0, or negative values of -5 and -3.
Therefore the Quadratic Equation for option D has no solution.
Answer:
x=2.125
y=0
C=19.125
Step-by-step explanation:
To solve this problem we can use a graphical method, we start first noticing the restrictions and , which restricts the solution to be in the positive quadrant. Then we plot the first restriction shown in purple, then we can plot the second one shown in the second plot in green.
The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.
So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.
since triangles are similar
4/x = 7x/7
=》 4/x = x
=》x^2 = 4
=》 x = 2
for right triangles a^2 = b^2+c^2
DF^2 = 7^2+2^2 = 53
=》 DF = 7.28
perimeter of DEF = 7 +2+7.28 = 16.28
450 centimeters because there are 100 centimeters in a meter.