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drek231 [11]
3 years ago
7

What is the midpoint of EC ?

Mathematics
1 answer:
sladkih [1.3K]3 years ago
4 0

<u>Given</u>:

Given that the graph OACE.

The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)

We need to determine the midpoint of EC.

<u>Midpoint of EC:</u>

The midpoint of EC can be determined using the formula,

Midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Substituting the coordinates E(2t,0) and C(2p, 2r), we get;

Midpoint=(\frac{2t+2p}{2},\frac{0+2r}{2})

Simplifying, we get;

Midpoint=(\frac{2(t+p)}{2},\frac{2r}{2})

Dividing, we get;

Midpoint=(t+p,r)

Thus, the midpoint of EC is (t + p, r)

Hence, Option A is the correct answer.

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We can use a formula to find the sum, or, since there's only three terms, we can find the value of each term then add up the results.

4 · (½)¹⁻¹ = 4

4 · (½)²⁻¹ = 2

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Step-by-step explanation:

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3 years ago
mrs. wilson placed the welcome mat shown below on her front door step. what is the area of her welcome mat in square feet?
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3 years ago
Area of a triangle with points at (-9,5), (6,10), and (2,-10)
Ann [662]
First we are going to draw the triangle using the given coordinates. 
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Distance from point A to point B:
d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}
d_{AB}= \sqrt{(6+9)^2+(10-5)^2}
d_{AB}= \sqrt{(15)^2+(5)^2}
d_{AB}= \sqrt{225+25}
d_{AB}= \sqrt{250}
d_{AB}=15.81

Distance from point A to point C:
d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}
d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}
d_{AC}= \sqrt{11^2+(-15)^2}
d_{AC}= \sqrt{121+225}
d_{AC}= \sqrt{346}
d_{AC}= 18.60

Distance from point B from point C
d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}
d_{BC}= \sqrt{(-4)^2+(-20)^2}
d_{BC}= \sqrt{16+400}
d_{BC}= \sqrt{416}
d_{BC}=20.40

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
s= \frac{AB+AC+BC}{2}
s= \frac{15.81+18.60+20.40}{2}
s= \frac{54.81}{2}
s=27.41

Finally, to find the area of our triangle, we are going to use Heron's formula:
A= \sqrt{s(s-AB)(s-AC)(s-BC)}
A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}
A= \sqrt{27.41(11.6)(8.81)(7.01)}
A=140.13

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0
3 years ago
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