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Anna007 [38]
2 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]2 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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CL 6-131. Solve each system using the method of your choice.
shepuryov [24]

System 1: The solution is (x, y) = (-4, 5)

System 2:  The solution is (x, y) = (\frac{11}{3}, -3)

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

2x + 3y = 7 ------ eqn 1

-3x - 5y = -13 --------- eqn 2

We can solve by elimination method

Multiply eqn 1 by 3

6x + 9y = 21 ------ eqn 3

Multiply eqn 2 by 2

-6x - 10y = -26 ------- eqn 4

Add eqn 3 and eqn 4

6x + 9y -6x - 10y = 21 - 26

-y = -5

y = 5

Substitute y = 5 in eqn 1

2x + 3(5) = 7

2x + 15 = 7

2x = -8

x = -4

Thus the solution is (x, y) = (-4, 5)

<h3><em><u>Second system of equation is:</u></em></h3>

8 - y = 3x ------ eqn 1

2y + 3x = 5 ----- eqn 2

We can solve by susbtitution method

From given,

y = 8 - 3x ----- eqn 3

Substitute eqn 3 in eqn 2

2(8 - 3x) + 3x = 5

16 - 6x + 3x = 5

3x = 16 - 5

3x = 11

x = \frac{11}{3}

Substitute the above value of x in eqn 3

y = 8 - 3x

y = 8 - 3 \times \frac{11}{3}\\\\y = 8 - 11\\\\y = -3

Thus the solution is (x, y) = (\frac{11}{3}, -3)

4 0
2 years ago
Minneapolis, Minnesota and Portland, Oregon have very similar latitudes. (Latitude measures how far north or south a place is, w
babymother [125]
A it was warmer in Minneapolis every day i think
4 0
2 years ago
Read 3 more answers
HELP PLEASE, ASAP!!!!
musickatia [10]

Answer:

A. step one get the lest common multiplier for x-2, x+2 ---->(x-2)

(x+2

B.-6

Step-by-step explanation:

8 0
1 year ago
Find the perimeter of a quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3). Round your answer to the ne
saveliy_v [14]

Answer: 20 units

Step-by-step explanation:

The distance between two points P(a,b) and Q(c,d) is given by :-

PQ=\sqrt{(d-b)^2+(c-a)^2}

Given : The vertices of a quadrilateral = C (−2, 1), D (2, 4), E (5, 0), and F (1, −3).

Distance between points C (−2, 1) and D (2, 4) :-

CD=\sqrt{(4-1)^2+(2-(-2))^2}\\\\\Rightarrow\ CD=\sqrt{(3)^2+(4)^2}\\\\\Rightarrow\ CD=\sqrt{9+16}=\sqrt{25}\\\\\Rightarrow\ CD=5\text{ units}

Distance between points D (2, 4) and E (5, 0) :-

DE=\sqrt{(4-0)^2+(2-5)^2}\\\\\Rightarrow\ DE=\sqrt{(4)^2+(-3)^2}\\\\\Rightarrow\ DE=\sqrt{16+9}=\sqrt{25}\\\\\Rightarrow\ DE=5\text{ units}

Distance between points E (5, 0) and F (1, −3).:-

EF=\sqrt{(-3-0)^2+(1-5)^2}\\\\\Rightarrow\ EF=\sqrt{(-3)^2+(-4)^2}\\\\\Rightarrow\ EF=\sqrt{9+16}=\sqrt{25}\\\\\Rightarrow\ EF=5\text{ units}

Distance between points C (−2, 1) and F (1, −3).:-

FC=\sqrt{(-3-1))^2+(1-(-2))^2}\\\\\Rightarrow\ FC=\sqrt{(-4)^2+(3)^2}\\\\\Rightarrow\ FC=\sqrt{16+9}=\sqrt{25}\\\\\Rightarrow\ FC=5\text{ units}

Now, the perimeter of the quadrilateral = CD+DE+EF+FC

=5+5+5+5=20\text{ units}

3 0
2 years ago
Read 2 more answers
Simon has a certain length of fencing to enclose a rectangular area. The function A models the rectangle's area (in square meter
ss7ja [257]

Answer:

The maximum area is 1,600 square meters

Step-by-step explanation:

<u><em>The complete question is</em></u>

What is the maximum area possible?

The given function area is modeled by A(w)=-w(w-80)

we know that

The given function is a vertical parabola open downward

The vertex is a maximum

The x-coordinate of the vertex represent the width for the maximum area

The y-coordinate of the vertex represent the maximum area

Convert the quadratic function in vertex form

A(w)=-w(w-80)\\\\A(w)=-w^{2}+80w

Factor -1

A(w)=-(w^{2}-80w)

Complete the square

A(w)=-(w^{2}-80w+1,600)+1,600

Rewrite as perfect squares

A(w)=-(w-40)^{2}+1,600 ----> function in vertex form

The vertex is the point (40,1,600)

therefore

The maximum area is 1,600 square meters

8 0
2 years ago
Read 2 more answers
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