The easiest way is to try the point (-4,1), that is, x=-4, y=1,
to see which equation works.
b works.
The usual way to do it is to find the equation of the circle
standard form of a circle is (x-h)²+(y-k)²=r², (h,k) are the coordinates of the center, r is the radius.
in this case, the center is (-2,1), so (x+2)²+(y-1)²=r²
the given point (-4,1) is for you to find r: (-4+2)²+(1-1)²=r², r=2
so the equation is (x+2)²+(y-1)²=2²
expand it: x²+4x+4+y²-2y+1=4
x²+y²+4x-2y+1=0, which is answer b.
Step-by-step explanation:
f(x) = 4x + 1
g(x) = x² - 5
(f+g)(x) = f(x) + g(x)
= (4x + 1) + (x² - 5)
= x² + 4x + (1 - 5)
= x² + 4x - 4
Option → C
Answer:
x = 2
y = —3
Step-by-step explanation:
3x — 9 = —2x + 1
x = 2
y = —3
<h3>Answer: A. 5/12, 25/12</h3>
============================
Work Shown:
12*sin(2pi/5*x)+10 = 16
12*sin(2pi/5*x) = 16-10
12*sin(2pi/5*x) = 6
sin(2pi/5*x) = 6/12
sin(2pi/5*x) = 0.5
2pi/5*x = arcsin(0.5)
2pi/5*x = pi/6+2pi*n or 2pi/5*x = 5pi/6+2pi*n
2/5*x = 1/6+2*n or 2/5*x = 5/6+2*n
x = (5/2)*(1/6+2*n) or x = (5/2)*(5/6+2*n)
x = 5/12+5n or x = 25/12+5n
these equations form the set of all solutions. The n is any integer.
--------
The two smallest positive solutions occur when n = 0, so,
x = 5/12+5n or x = 25/12+5n
x = 5/12+5*0 or x = 25/12+5*0
x = 5/12 or x = 25/12
--------
Plugging either x value into the expression 12*sin(2pi/5*x)+10 should yield 16, which would confirm the two answers.
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701
