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Nuetrik [128]
3 years ago
5

Can someone please help me on this one

Mathematics
2 answers:
BabaBlast [244]3 years ago
6 0

Answer:

4. -6

5. 4

6. 1

7. -2

8. -5

9. 6

Aleonysh [2.5K]3 years ago
4 0

Answer:

4. -6, 5. 4, 6. 1 7. -2 8. -5 9. 6

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PLEASE HELP ME........​
LenKa [72]

Answer:

d

Step-by-step explanation:

4 0
3 years ago
What is the domain of a function?
Katen [24]
B . The set of all first elements of the function is the answer
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3 years ago
In the following question solve for X please help!
Mariulka [41]

Answer:

<u>x = 6</u>

Step-by-step explanation:

Taking the sides in proportion :

  • EB/DC = AB/AC
  • 8/24 = -3 + 2x/27
  • 1/3 = -3 + 2x/27
  • 3(-3 + 2x) = 27
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8 0
2 years ago
Read 2 more answers
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
3 years ago
Do you know parallelogram WXYZ Parkway 270° around point W what will be the length of the image WZ
Lerok [7]

Answer:

5 units

Step-by-step explanation:

Even if there is a transformation, it is asking for the length.

Therefore the length of WZ is 5 units

7 0
2 years ago
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