Answer:
p = (2k + 1)i + j
q = 25i + (k + 4)j
p . q = 11
Find the dot product of p and q
(2k + 1)(25) + (k + 4) = 11
50k + 25 + k + 4 = 11
51k = -18
k = - 18/51 = –6/17.✅
b) p.q = |p||q|cos∅
p= (2k + 1)i + j = [2(-6/17) + 1)i + j ]= 5/17i + j
|p| = √ (5/17)² + 1²
|p| = √314/ 17 ( The square root doesn't cover the 17) = 1.0424
q = 25i + (k + 4)j = 25i + ( -6/17 + 4)j
q = 25i + 62/17 j
|q| = √ 25² + (62/17)²
= 25.2646
Now applying them to the formula above
We already have p.q = 11 from the question.
11 = (1.0424)(25.2646)Cos∅
11 = 26.3358Cos∅
Cos∅ = 11/26.3358
Cos∅ = 0.4177
∅ = 65.31°.
This should be it.
Hope it helps
Answer:
girls = 125
boys = 87
Step-by-step explanation:
Total number of students in the 8th grade class = 212
Let g represent girls and b represent boys
Number of girls, g = 2b -49
Note that boys and girls in the class give a total of 212
So,
g + b = 212 ....eq 1
Slot in the value of g in the equation
2b - 49 + b = 212
2b + b - 49 = 212
3b - 49 = 212
Add 49 to both sides
3b - 49 + 49 = 212 + 49
3b = 261
Divide both sides by the coefficient of b which is 3
b = 87
Now that we've found the number of boys, let's find that of girls
We will slot in the value of b in eq 1
g + 87 = 212
Subtract 87 from both sides
g + 87 - 87 = 212 - 87
g = 125
So the number of boys is 87 and the number of girls is 125
Parameterize the part of the surface we care about, denoted
, by

with
and
.
Then

The area is given by the surface integral


Answer:
8.0
Step-by-step explanation:
Using the SOH CAH TOA identity
cos theta = adjacent/hypotenuse
cos 37 = x/10
x = 10cos 37
x = 7.99
The vale to the nearest tenth is 8.0