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3 years ago

Which equation correctly applies the distributive property?

2 answers:

7 0

<span>distributive property

</span><span>D.(0.5 ⋅ 0.5) + (0.5 ⋅ 0.3) + (0.5 ⋅ 0.2) = 0.5 ⋅ (0.5 + 0.3 + 0.2) </span><span>

</span>

nika2105 [10]3 years ago

4 0

distributive property

D.(0.5 ⋅ 0.5) + (0.5 ⋅ 0.3) + (0.5 ⋅ 0.2) = 0.5 ⋅ (0.5 + 0.3 + 0.2)

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PLEASE WILL GIVE 50 POINTS PLEASE ANSWER

leva [86]

Answer:

AB = 75

BC = 60

AC = 45

m∠A = 53°

m∠B = 37°

m∠C = 90°

Step-by-step explanation:

<u>Trigonometric ratios</u>

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Given:

$\sf \tan(A)=\dfrac{60}{45}$

Therefore:

side opposite angle A = BC = 60
side adjacent angle A = AC = 45

To find the length of AB (the hypotenuse), use Pythagoras’ Theorem:

$a^2+b^2=c^2$

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

⇒ AC² + BC² = AB²

⇒ 45² + 60² = AB²

⇒ AB² = 5625

⇒ AB = √5625

⇒ AB = 75

To find m∠A:

$\implies\sf \tan(A)=\dfrac{60}{45}$

$\implies\sf A=\tan^{-1}\left(\dfrac{60}{45}\right)$

$\implies\sf A=53^{\circ}\:(nearest\:degree)$

m∠C = 90° (as it is a right angle)

The interior angles of a triangle sum to 180°

⇒ m∠A + m∠B + m∠C = 180°

⇒ 53° + m∠B + 90° = 180°

⇒ m∠B = 180° - 53° - 90°

⇒ m∠B = 37°

4 0

2 years ago

Cylinder a has a mass of 3 kg and cylinder b has a mass of 8 kg. Determine the speed of a after it has moved 2 m starting from r

german

The rate of change of an object's velocity with respect to time is defined as acceleration. The speed of a after it has moved 2m starting from rest is 1.98 m/s².

<h3>What is acceleration?</h3>

The rate of change of an object's velocity with respect to time is defined as acceleration. Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.

We can see that the acceleration of 3kg mass is double that of the 8kg mass. Therefore, Writing the force conservation equations altogether,

(T - 3g)/3 = 2(8g - 2T)/8

T= 3.6*g

Now, the Acceleration of 8kg mass can be written as

a = (8g - 2T)/8

= 0.8g/8

=0.1g

Therefore, acceleration of mass 3kg = 2×a = 0.2g

Further, as per the law of motion we can write,

v² - u² = 2as

Substituting the values, u =0; a=0.1g; s=2m; we will get,

v=1.98m/s

Hence, the speed of a after it has moved 2m starting from rest is 1.98 m/s².