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3 years ago

First question. The midpoint of SR is...

Mathematics

1 answer:

Blizzard [7]3 years ago

6 0

The formula of a midpoint SR:

$M\left(\dfrac{x_S+x_R}{2},\ \dfrac{y_S+y_R}{2}\right)$

We have

S(4, 1) and M(7, -5)

Substitute:

$\dfrac{4+x_R}{2}=7\qquad|\cdot2\\\\4+x_R=14\qquad|-4\\\\x_R=10\\\\\dfrac{1+y_R}{2}=-5\qquad|\cdot2\\\\1+y_R=-10\qquad|-1\\\\y_R=-11$

<h3>Answer: R(10, -11)</h3>

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WILL MARK AS BRAINLIEST PLEASE HELP!!!

neonofarm [45]

Step-by-step explanation:

2x = y-10

Rewrite as: 2x + 10 = y

Therefore:

2x+10 = y

2x +7 = 2y

Subtract the equations:

2x + 10 = y

- 2x + 7 = 2y

______________

3 = -y

Therefore y = -3

Substiture y = -3 into the second equation:

2x + 7 = 2(-3)

2x + 7 = -6

2x = -13

x= -6.5

Answer : (-6.5, -3)

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3 years ago

Read 2 more answers

the measures of the side of a triangle are 4√2 cm, 4√2 cm qnd 8√2 cm. what is the perimeter of the triangle?

spin [16.1K]

Answer:

$16\sqrt{2}$

Step-by-step explanation:

Add up all the sides to get the perimeter:

$4\sqrt{2} + 4\sqrt{2} +8\sqrt{2} = 16\sqrt{2}$

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3 years ago

The graph below represents which system of inequalities?

attashe74 [19]

Answer:

y ≤ 2x -2 Option C.

Step-by-step explanation:

Equation of the dotted line which passes through (3,0) and (0,8) will be

y = mx+C

m = $\frac{y-y'}{x-x'}=\frac{3-0}{0-3}$ = -1

and C = 3

So equation will be y = x + 3

But shaded blue part is above the dotted line so inequality will be y> -x +3

Equation of the second solid line which passes through ( 0,-2) and (1,0) will be

y = mx + C

m = $\frac{y-y'}{x-x'}$ = $\frac{0+2}{1-0}$ = 2

and C = -2

So equation of the line will be y = 2x - 2

But shaded yellow part is below the solid line.

So inequality will be

y ≤ 2x -2

Option C. is the answer.

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3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

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