Answer:
Step-by-step explanation:
9+10=19.
Replace the 0 in the ten with the 9.
Option A

<em><u>Solution:</u></em>
Given that we have to rewrite with only sin x and cos x
Given is cos 3x

We know that,

Therefore,
---- eqn 1
We know that,


Substituting these values in eqn 1
-------- eqn 2
We know that,

Applying this in above eqn 2, we get



Therefore,

Option A is correct
Answer:
5. -18x^3y^4 +54x^2y^4
6. 6x^2 -40x +50
Step-by-step explanation:
Use the distributive property. The factor outside parentheses multiplies each term inside parentheses.
__
<h3>5.</h3>

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<h3>6.</h3>

Answer:
ρ_air = 0.15544 kg/m^3
Step-by-step explanation:
Solution:-
- The deflated ball ( no air ) initially weighs:
m1 = 0.615 kg
- The air is pumped into the ball and weight again. The new reading of the ball's weight is:
m2 = 0.624 kg
- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.
m_air = Δm = m2 - m1
m_air = 0.624 - 0.615
m_air = 0.009 kg
- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:
V_air = 4*π*r^3 / 3
V_air = 4*π*0.24^3 / 3
V_air = 0.05790 m^3
- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:
ρ_air = m_air / V_air
ρ_air = 0.009 / 0.05790
Answer: ρ_air = 0.15544 kg/m^3