The weight of the new student is 27 kg.
Average weight
= total weight ÷total number of students
<h3>
1) Define variables</h3>
Let the total weight of the 35 students be y kg and the weight of the new student be x kg.
<h3>2) Find the total weight of the 35 students</h3>
<u>
</u>
y= 35(45)
y= 1575 kg
<h3>3) Write an expression for average weight of students after the addition of the new student</h3>
New total number of students
= 35 +1
= 36
Total weight
= total weight of 35 students +weight of new students
= y +x
<h3>4) Substitute the value of y</h3>
<h3>5) Solve for x</h3>
36(44.5)= 1575 +x
1602= x +1575
<em>Subtract 1575 from both sides:</em>
x= 1602 -1575
x= 27
Thus, the weight of the new student is 27 kg.
Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)
If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.
On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.
For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.
Then the range of F(x) is the interval [-π, π].
Answer:
<h3>x = -3</h3>
Step-by-step explanation:
First let us get the equation of the coordinates
y-y0 = m(x-x0)
Using the coordinates ( - 3, 2 ), ( - 1, 0 )
m = 0-2/-1-(-3)
m = -2/2
m = -1
Substitute m = -1 and (-1, 0) into the formula
y - 0 = -1(x+1)
y = -x-1
f(x) = -x-1
Since f(x) = 2
2 = -x-1
-x = 2+1
-x = 3
x = -3
Hence the value of x is -3
