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Inessa [10]
3 years ago
8

Indicate the equation of the given line in standard form.The line containing the median of the trapezoid whose vertices are R(-1

, 5) , S(1, 8), T(7, -2), and U(2, 0).
Mathematics
1 answer:
maxonik [38]3 years ago
7 0
RS => y - 5 = (8 - 5)/(1 - (-1)) (x - (-1))
y - 5 = 3/2 (x + 1) => slope = 3/2
ST => y - 8 = (-2 - 8)/(7 - 1) (x - 1)
y - 8 = -10/6 (x - 1) = -5/3 (x - 1) => slope = -5/3
TU => y - (-2) = (0 - (-2))/(2 - 7) (x - 7)
y + 2 = 2/5(x - 7) => slope = 2/5
UR => y = 5/(-1 - 2) (x - 2)
y = -5/3 (x - 2) => slope = -5/3

The median is the line joining the midpoints of the non-parallel sides.
Midpoint of RS = ((-1 + 1)/2, (5 + 8)/2) = (0, 13/2)
Midpoint of TU = ((7 + 2)/2, -2/2) = (9/2, -1)

Equation of the line joining (0, 13/2) and (9/2, -1) is given by y - 13/2 = (-1 - 13/2)/(9/2) x
y - 13/2 = (-15/2)/(9/2) x
y - 13/2 = -15/9x
18y - 117 = -30x
30x + 18y = 117
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Yes MNO is similar to PQO

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3 years ago
Any help plzzz I’ll appreciate it so much thank u
docker41 [41]

9514 1404 393

Answer:

  5√5

Step-by-step explanation:

Use the distance formula:

  d = √((x2 -x1)² +(y2 -y1)²)

  d = √((-7-(-2))² +(-7-3)²) = √((-5)² +(-10)²) = √(25 +100)

  d = √125 = √(25·5)

  d = 5√5 . . . . distance between the points

3 0
3 years ago
I need help fast i don’t understand please give detailed explanation
alex41 [277]

Answer:

<4 = 63°

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5 0
3 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
2 years ago
How to find a slope of a line
max2010maxim [7]

Answer:

The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points

8 0
2 years ago
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