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3 years ago

If medain of 10, 20, x+5,x+10,50,60 is 35 then find x

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

X = 27

Step-by-step explanation:

10, 20, 32, 37, 50, 60

First things first, take off 10, 20, 50, and 60.

Now, you remain with 32 and 37. Now from here it is simple.

Average of 32 and 37 = 34.83

Round 34.83 to 35

32 averaging 37 (Rounding Terms) ≈ 35

So X is 27

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The distance between the two points (4,2) and (8,k) is 5 units. Find value(s)of k

Sliva [168]

D^2 = (x2 - x1)^2 + (y2 - y1)^2
d^2 = (8 - 4)^2 + (k - 2)^2
d = 5 so
5^2 = 4^2 + k^2 - 4k + 4
25 = 16 + k^2 - 4k + 4
25 = 20 + k^2 - 4k
0 = k^2 - 4k - 5
0 = (k - 5)(k + 1)
k has 2 answers
k - 5 = 0
k = 5
k + 1 = 0
k = - 1

The two answers are
(8, 5) and (8, -1) <<<< answer

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3 years ago

2(q–8)< – 10 solve for q

natka813 [3]

Answer:

$= q < 3$

Step-by-step explanation:

$2(q - 8) < - 10 \\ 2q - 16 < - 10 \\ 2q < - 10 + 16 \\ 2q < 6 \\ \frac{2q}{2} < \frac{6}{2} \\ q < 3$

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2 years ago

Rene is going to the lake to visit some friends. If the lake is 60 miles away and Rene is driving at 40 miles per hour the entir

MArishka [77]

It will take her an hour and a hlaf

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3 years ago

Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are neither parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are parallel

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are Neither parallel nor perpendicular

Neither

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are perpendicular

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are parallel

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are neither parallel nor perpendicular

Neither

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

Neither

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is Neither

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The graph below shows the path of a model rocket after its launch from the ground.

Zarrin [17]

The answer the third one

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2 years ago