3³ - (4)(2) find da value or die mwaha

Which of following not a way to represent the solution of inequality 5(x + 2) GREATER THAN OR EQUAL TO 7x + 2(x − 1)?

mixer [17]

So because it says NOT a way....I think it's A...Maybe I'm wrong but whatever.

6 0

3 years ago

Read 2 more answers

2. Katryn swims at a rate of 50 meters in 40 seconds. At this rate, how many meters

Sunny_sXe [5.5K]

Answer:

D.) 75 meters

Step-by-step explanation:

1 minute is 60 seconds

So, you divide 60 divided by 40=1.5

1.5 times 50=75

75 meters

D.)

Hope this helped!! :)

Brainliest?!?!

Stay safe and have a nice day/night/afternoon!!!

4 0

3 years ago

Find the area of the regular polygon below and round your answer to the nearest tenth.

stira [4]

The 9-sided polygon will have a central angle between the vertex and the center of one side of 360°/(2*9) = 20°. Then the apothem is
a = (6 in/2)*cot(20°) ≈ 8.242432 in
The perimeter is 9*6 in = 54 in, so the area is
A = (1/2)Pa = (1/2)*(54 in)*(8.242432 in) ≈ 222.5 in²

4 0

3 years ago

Let P be the parabola with focus (0,4) and directrix y=x.Write an equation whose graph is a parabola with a vertical directrix t

prisoha [69]

Answer:

y²=4√2.x

Step-by-step explanation:

The focus is at (0,4) and directrix is y=x or x-y =0, for a parabola P.

The distance between the focus and the directrix of the parabola P is

$\frac{ |0-4| }{\sqrt{(1)^{2}+(-1)^{2} } }$ = $\frac{4}{\sqrt{2} }$

{Since the perpendicular distance of a point (x1, y1) from the straight line ax+by+c =0 is given by $\frac{ |ax1+by1+c| }{\sqrt{a^{2}+b^{2} } }$ }

Let us assume that the equation of the parabola which is congruent with parabola P is y²=4ax

{Since the parabola has vertical directrix}

Hence, the distance between focus and the directrix is 2a = $\frac{4}{\sqrt{2} }$ , {Two parabolas are congruent when the distances between their focus and the directrix are same}

⇒ a=√2

Therefore, the equation of the parabola is y²=4√2.x (Answer)

3 0

3 years ago

Select ALL the correct answers.

Marina CMI [18]

Answer:

The median of A is the same as the median of B.

The interquartile range of B is greater than the interquartile range of A.

Step-by-step explanation:

Given that:

A = number of runs allowed in first 9 games

A = {1, 4, 2, 2, 3, 1, 1, 2, 1}

Rearranging A : 1, 1, 1, 1, 2, 2, 2, 3, 4

Median A = 1/2(n + 1) th term

Median A = 1/2(10) = 5th term = 2

Q1 of A = 1/4(10) = 2.5th term = (1 + 1)/ 2 = 1

Q3 of A = 3/4(10) = 7.5th term = (2+3)/2 = 2.5

Interquartile range = Q3 - Q1 = 2.5 - 1 = 1.5

Number of runs allowed in 10th game = 9

B = {1, 4, 2, 2, 3, 1, 1, 2, 1, 9}

Rearranging B = 1, 1, 1, 1, 2, 2, 2, 3, 4, 9

Median A = 1/2(n + 1) th term

Median A = 1/2(11) = 5.5th term = (2+2)/2 = 2

Q1 of A = 1/4(11) = 2.75th tetm = (1 + 1)/ 2 = 1

Q3 of A = 3/4(11) = 8.25th term = (3+4)/2 = 3.5

Interquartile range = Q3 - Q1 = 3.5 - 1 = 2.5

Median A = 2 ; median B = 2

IQR B = 2.5 ; IQR A = 1.5 ; IQR B > IQR A

7 0

2 years ago