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antoniya [11.8K]
2 years ago
7

PQ has a length of 17 units with P(-4,7). If the x- and y-coordinates of Q are both greater than the x- and y-coordinates of P,

what are possible integer value coordinates of
Q? Explain
Let PQ be the hypotenuse of a right triangle that also has a horizontal leg and a vertical leg. The hypotenuse then has length 17, and a Pythagorean triple can be used to say the shorter leg has length___ and the longer leg has length____If the shorter leg is horizontal, then Q is described by the ordered pair___If the shorter leg is vertical,
then Q is described by the ordered pair____ FILL IN THE BLANK PLS
Mathematics
1 answer:
Paladinen [302]2 years ago
7 0

The complete paragraph on Pythagorean theorem is shown below:

<em>Let PQ be the hypotenuse of a right triangle that also has a horizontal leg and a vertical leg. The hypotenuse then has length 17, adn a Pythagorean triple can be used to say the shorter leg has length 4 and the longer leg has length 22. If the shorter leg is horizontal, then Q is described by the ordered pair (x, y) = (4, 22). If the shorter leg is vertical is vertical, then Q is described by no ordered pairs.</em>

<h3>What are the coordinates of the missing end of a line segment?</h3>

Given the coordinates of the two end we can determine the length of a line segment by using the Pythagorean theorem in the following form and considering that the coordinates of point Q are integers greater than their counterparts of point P:

17 = √[[x - (-4)]² + (y - 7)²]

[x - (-4)]² + (y - 7)² = 289

(x + 4)² + (y - 7)² = 289

Then, we proceed to graph the circle and we find that the point satisfying the conditions indicated are (x, y) = (4, 22).

To learn more on Pythagorean theorem: brainly.com/question/14930619

#SPJ1

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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}}

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3 years ago
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Answer: A

Step-by-step explanation:

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Y_Kistochka [10]

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Step-by-step explanation:

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What best describes the end behavior of the graph of f(x)=-x^4+5x-5
Neko [114]

Answer:

It describes multiplication

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8 0
3 years ago
Two fair dice are rolled 3 times and the sum of the numbers that come up is recorded. Find the probability of these events. a. T
alexandr402 [8]

Answer:

Step-by-step explanation:

Given that two fair dice are rolled 3 times and the sum of the numbers that come up is recorded.

a) The sum is 4 on each of the three rolls.

(1,3) (2,2) (3,1) HEnce for one throw prob = 3/36 =1/12

Prob for getting same sum 4 in 3 rolls = (\frac{1}{12} )^3

(since each trial is independent)

b) Prob sum =4 in exactly 2 rolls

= 3C2 (1/12)^2 (11/12)=\frac{33}{12^3}

5 0
3 years ago
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