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expeople1 [14]
3 years ago
14

Find the coordinates of the missing endpoint if P is the midpoint of Line Segment NQ.

Mathematics
1 answer:
dimaraw [331]3 years ago
8 0
  • Midpoint formula is (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) .
<h3>19.</h3>

So starting with this one, we will be solving for the coordinates of the unknown endpoint separately. Starting with the x-coordinate, since we know that the midpoint x-coordinate is 5 and the x-coordinate of N is 2, our equation is set up as such: \frac{x+2}{2}=5 From here we can solve for the x-coordinate of Q.

Firstly, multiply both sides by 2: x+2=10

Next, subtract both sides by 2 and your x-coordinate is x=8

With finding the y-coordinate, it's a similar process as with the x-coordinate except that we are using the y-coordinates of the midpoint and endpoint N.

\frac{y+0}{2}=2\\ y=4

<u>Putting it together, the missing endpoint is (8,4).</u>

<em>(The process is pretty much the same with the other problems, so I'll go through them real quickly.)</em>

<h3>20.</h3>

\frac{x+5}{2}=6\\ x+5=12\\ x=7

\frac{y+4}{2}=3\\ y+4=6\\ y=2

<u>The missing endpoint is (7,2).</u>

<h3>21.</h3>

\frac{x+3}{2}=-1\\ x+3=-2\\ x=-5

\frac{y+9}{2}=5\\ y+9=10\\y=1

<u>The missing endpoint is (-5,1).</u>

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Answer: 10.3 would be a mixed number

Step-by-step explanation:

Divide 12 into 46. You get 3 with a remainder of 10. 3 is the whole number of the mixed number, 10 is the numerator of the fraction, and 12 is the denominator. The fraction can be reduced to 5/6.

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Solve for X. (1 point)<br> -ax + 3b &gt; 5
Anarel [89]

Answer:

x<(3b-5)/a

Step-by-step explanation:

Subtract 3b from both sides: -ax>5-3b

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3 years ago
If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi
Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

BCPO is a square

∴ BC = OP = CP = OB = 210mm

Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

(Hypotenuse)² = Sum of square of the other two sides

(OC)² = (OB)² + (BC)²

(OC)² = 210² + 210²

(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

An isosceless tringle is a triangle that has two equal sides

Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴  ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}

Note that DP = 297 - 210 = 87mm

∠PDO + ∠DOP + 90 = 180

∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

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