Me im struggling so bad will give brainliest type the correct answer in the box. if necessary, use / for the fraction bar. in δpqr, ∠p and ∠q are complementary angles. if sin q= 4/5 , cos p + cos q= ? ?

Remark
If the two angles are complimentary they add up to 90°. So the first job is to find the third angle. In the diagram below, the third angle is labeled C, so you must find it first. 

After that, you need to use the Pythagorean Theorem to solve for QC which at this point is not known. 

After completing that, You need to find the cosine of P and Q so that they can be added together.

Step One
Find angle C
All triangles have 180°
P + Q are complementary.      Given
P + Q = 90°                             Definition
P + Q + C = 180                      Property of a triangle.
90 + C = 180                           Subtract 90 from both sides.
C = 180 - 90
C = 90°

Step Two
Find QC
When you have a right triangle the Pythagorean Theorem can be used to find any missing side.
Sin(Q) = opposite / hypotenuse  [See diagram]
Sin(Q) = 4/5                                Given

a = 4
b = ??
c = 5

4^2 + b^2 = 5^2
16 + b^2 = 25  Subtract 16 from both sides.
b^2 = 25 - 16
b^2 = 9
Take the square root of both sides.
sqrt(b^2) = sqrt(9)
b = 3

Step 3
Define Cos(Q) and Cos(P)
Cos(Q) = adjacent side (3) over the hypotenuse (5)
Cos(Q) = 3/5

Cos(P) has the same basic definition. The adjacent side is 4 for angle P
Cos(P) = 4/5

Step Four
Add the two findings from Step 3

Cos(Q) + Cos(P) = 3/5 + 4/5 = 7/5 <<<<< Answer

The temperature of the second engine increases more for each increase of 1 cycle per second.

The second engine will be hotter.

Step-by-step explanation:

Each engine's rotation got faster, its temperature increased at a constant rate.  

The first engine's temperature (in degrees Celsius) as a function of its rotation speed (in cycles per second) is given by the table attached.

So, as the rate of change of temperature with increase in speed of rotation is linear, so, we can express the relation as  

T = aS + b (1)

Now, from the table we can write 26.4 = 8a + b (2)

and 28.8 = 11a + b (3)

Now, from equation (2) and (3), we get

11a - 8a = 2.4

⇒ 3a = 2.4  

⇒ a = 0.8  

Again, from equation (2) we get,

26.4 = 8 × 0.8 + b

⇒ b = 20

Therefore, the equation (1) becomes T = 0.8S + 20 (4)

Now, from the graph of the second engine, the y-intercept is 22 and slope of the straight line is {Since, the graph passes through the points (0,22) and (5,28).

Therefore, the equation can be written as  

T = 1.2S + 22 (5)

Therefore, the temperature of the second engine increases more for each increase of 1 cycle per second as from equations (4) and (5) the rate of change of temperature with respect to rotational speed is 1.2 for the second engine which is more than 0.8 for first engine. (Answer)

Now, for S = 20 cycles per second, from equation (4) we get,

T = 16 + 20 = 36° C

And from equation (5) we get, T = 24 + 22 = 46° C

So, the second engine will be hotter. (Answer)