Which explanation correctly tells whether these ratios are equivalent and why?

Mathematics

1 answer:

Reptile [31]2 years ago

6 0

<h3><u>Answer:</u></h3>

<u>Option A (Yes because 3 x 2 = 6 and 14 x 2 = 28.) is correct.</u>

<h3><u>Step-by-step explanation:</u></h3>

<u>This is because if we simplify 3/6 and 14/28, both of them will get 1/2. This means that only A and C are possible.</u>

=> Option A is correct because 3 x 2 does equal 6 and 14 x 2 does equal 28.
=> Option C is incorrect because 3 x 5 does not equal 14 and 6 x 5 does not equal 28.

Hoped this helped.

$BrainiacUser1357$

You might be interested in

A 100 ft. ladder rests on top of a hook and ladder truck with its base 11 feet from the ground. When the angle of elevation of t

vichka [17]

100 plus 11 co sin 81

6 0

2 years ago

A triangle has angle measures of 30 degrees and 70 degrees. What is the measure of the third angle?

cricket20 [7]

Because a triangle's angles need to equal 180, you can subtract 100(30+70) and you will get 80. So 80 will be your third angle.

8 0

3 years ago

A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt

Katena32 [7]

Answer:

$2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}$

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be $y$ . The magnitude of $40\text{ m/s}$ will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is $60^{\circ}$ , we have:

$\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}$ (Recall that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ )

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation $v_f^2=v_i^2+2a\Delta y$ to solve this problem, where $v_f/v_i$ is final and initial velocity, respectively, $a$ is acceleration, and $\Delta y$ is distance travelled. The only acceleration is acceleration due to gravity, which is approximately $9.8\:\mathrm{m/s^2}$ . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.