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2 years ago

6

Find the slope of the line

2 answers:

jeka57 [31]2 years ago

6 0

6/3 or 2/1 up 3 over 1

antoniya [11.8K]2 years ago

4 0

Y= $\frac{2}{3}$ x

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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.

What we know:

$v_i=20\sqrt{3}\text{ m/s}$
$a=-9.8\:\mathrm{m/s^2}$
$\Delta y =50\text{ m}$

Solving for $v_f$ :

$v_f^2=(20\sqrt{3})^2+2(-9.8)(50),\\v_f^2=1200-980,\\v_f^2=220,\\v_f=\sqrt{220}=\boxed{2\sqrt{55}\text{ m/s}}$

3 0

2 years ago

Identify THREE vocabulary words that also refer to x-intercepts.

Marianna [84]

Answer:

X-intercepts are also called zeros, roots, solutions, or solution sets.

7 0

2 years ago

Read 2 more answers

It is possible none of the choices is the solution?

velikii [3]

Answer:

I am pretty sure the answer would be B!

Hope this helps

Mark me brainliest if I'm right :)

6 0

2 years ago

What is the slope of the line that passes through the pair of points (1/3, -1) and ( -2, 9/2)

UkoKoshka [18]

Answer:

$\boxed{m = - \frac{33}{14} }$

.

Step-by-step explanation:

Use the form below

$\boxed{\boxed{m = \frac{y_2 - y_1}{x_2 - x_1} }}$

Where

$m$ is a slope
$(x_1,~y_1)$ and $(x_2,~y_2)$ are the point of the line

.

So, the slope is

$(\frac{1}{3} ,~ -1) \to x_1 = \frac{1}{3} ~and~y_1 = -1$

$(-2,~ \frac{9}{2} ) \to x_2 = -2~and~y_2 = \frac{9}{2}$

.

$m = \frac{\frac{9}{2} -(-1)}{-2-\frac{1}{3} }$

$m = \frac{\frac{11}{2} }{\frac{-7}{3} }$

$m = \frac{11}{2} \div (-\frac{7}{3} )$

$m = \frac{11}{2} \times (-\frac{3}{7} )$

$m = - \frac{33}{14}$

.

Happy to help:)

7 0

2 years ago

Read 2 more answers

The mean weight 20 students in a room is 65 kg one student one student who weight is 88kg went into the room what is the new mea

umka2103 [35]

The new mean is 66.09

<u>Explanation:</u>

Mean weight of 20 students = 65kg

One student with 88kg entered the room.

New mean = ?

$\frac{x_1+x_2+....+x_2_0}{20} = 65$

$\frac{x_1+x_2+....+x_2_0 + 88}{21} = ?$

The above equation can further be written as:

$x_1 + x_2 +....+x_2_0 = 65 X 20\\\\x_1 + x_2 +....+x_2_0 = 1300$

New mean = $\frac{1300 + 88}{21}$

= 66.09

Therefore, the new mean is 66.09

6 0

2 years ago