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Anna007 [38]
4 years ago
15

Use a coordinate proof to prove that mid segment MN of triangle PQR is parallel to PR and half the length of PR. Which is the fi

rst best step?
A. Place the triangle on a coordinate grid such that vertex P is at the origin, and PR lies on x-axis.


B. place the triangle on a coordinate grid such that vertex P is on y-axis and vertex R us on the x-axis


C. Place the triangle on a coordinate grid such that vertex Q is at the origin


D. Place the triangle on a coordinate grid such that QR lies on the x-axis and PQ lies in the y-axis

Mathematics
1 answer:
IRINA_888 [86]4 years ago
4 0

Answer:

D. Place the triangle on a coordinate grid such that QR lies on the x-axis and PQ lies in the y-axis

Step-by-step explanation:

A midsegment of the triangle is the line segment the connects the midpoints of two sides of the triangle. This midsegmet is parallel to the base.

Constructing a triangle with two sides over the x-axis and y-axis respectively makes it easier to verify that the midsegment is half the base.

So, it's D. The best option.

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What fraction of 9 is 3​
Alex Ar [27]

Answer:

1/3

Step-by-step explanation:

Take the part over the whole

3/9

We can simplify

Divide the top and bottom by 3

1/3

4 0
3 years ago
Read 2 more answers
How many real solutions does the equation 8x^2 − 10x + 15 = 0 have?
Leto [7]

Answer:

It has no real solutions.

Step-by-step explanation

Once you calculate the discriminant,  D= (-10)^2 -4 (8)(15), you simplify the expression to get D= -380, which has no real solutions.

6 0
4 years ago
Read 2 more answers
What is the missing statement in the proof? Scroll down to see the entire proof.
Eva8 [605]

Answer: C. \angle BCA\cong \angle DCB is the missing statement.

Explanation: Given: ΔABC with m∠ABC = 90° (view diagram)

Prove: AB^2 + BC^2 = AC^2

Proof:  1. Draw BD\perp AC (construction)

 2. \angle ABC\cong\angle BDC  (Angles with the same measure are congruent.)

3. \angle BCA\cong \angle DCB (Reflexive Property of Congruence)

4. \triangle ABC\sim \triangle BDC  (AA criterion for similarity)

5. BC:DC=AC:BC  (Corresponding sides of similar triangles are proportional.)

 6.BC^2= AC\times DC  (cross multiplication)----------(1)

7. \angle ABC\cong \angle ADB (Angles with the same measure are congruent)

8. \angle BAC\cong \angle DAB (Reflexive Property of Congruence)

9. \triangle ABC\sim \triangle ADB  (AA criterion for similarity)

10.  AB:AD=AC:AB⇒AB^2=AC.AD --------------(2)

equation (1) + equation (2) ⇒AB^2+BC^2= AC(DC+AD)⇒AB^2+BC^2= AC^2


6 0
3 years ago
Read 2 more answers
What is the solution of the system of equations?
umka21 [38]

No solution of the system of equations y = -2x + 5 and -5y = 10x + 20 ⇒ 2nd answer

Step-by-step explanation:

Let us revise the types of solutions of a system of linear equations

  • One solution
  • No solution when the coefficients of x and y in the two equations are equal and the numerical terms are different
  • Infinitely many solutions when the coefficients of x , y and the numerical terms are equal in the two equations

∵ y = -2x + 5

- Add 2x to both sides

∴ 2x + y = 5 ⇒ (1)

∵ -5y = 10x + 20

- Subtract 10x from both sides

∴ -10x - 5y = 20

- Divide both sides by -5

∴ 2x + y = -4 ⇒ (2)

∵ The coefficient of x in equation (1) is 2

∵ The coefficient of x in equation (2) is 2

∴ The coefficients of x in the two equations are equal

∵ The coefficient of y in equation (1) is 1

∵ The coefficient of y in equation (2) is 1

∴ The coefficients of y in the two equations are equal

∵ The numerical term in equation (1) is 5

∵ The numerical term in equation (2) is -4

∴ The numerical terms are different

From the 2nd rule above

∴ No solution of the system of equations

No solution of the system of equations y = -2x + 5 and -5y = 10x + 20

Learn more:

You can learn more about the system of equations in brainly.com/question/6075514

#LearnwithBrainly

7 0
4 years ago
78362 divided by 509
vaieri [72.5K]

Answer:

153.952848723

djjxhdndhdududjdn

5 0
3 years ago
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