Plz help me with this

What is the factored form of -1/2(32x - 40) - (20x - 4)?

inna [77]

Answer:

12(-3x + 2)

Step-by-step explanation:

-1/2 ( 32x - 40) - (20x - 4) = -16x + 20 - 20x + 4 = -36x + 24 = 6(-6x + 4) = 12(-3x + 2)

8 0

3 years ago

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A theater sells adult tickets for $8.00 and Children's tickets for 5.50. Sally bought some of each kind (not the same number) an

sdas [7]

8x +(5.5)y=78.50 is the equation.

8 0

3 years ago

Felipa says she has an easy way to estimate the sales tax when she makes a purchase. The sales tax in her city is 9.05%. She kno

Bess [88]

Answer:

a) 1 / 10

b) $15/10 = $1.50

Step-by-step explanation:

8 0

3 years ago

I have an assignment and I am having trouble with it. Can someone please help ASAP???

bezimeni [28]

Answer:

A) Find the sketch in attachment.

In the sketch, we have plotted:

- The length of the arena on the x-axis (90 feet)

- The width of the arena on the y-axis (95 feet)

- The position of the robot at t = 2 sec (10,30) and its position at t = 8 sec (40,75)

The origin (0,0) is the southweast corner of the arena. The system of inequalities to descibe the region of the arena is:

$0\leq x \leq 90\\0\leq y \leq 95$

B)

Since the speed of the robot is constant, it covers equal distances (both in the x- and y- axis) in the same time.

Let's look at the x-axis: the robot has covered 10 ft in 2 s and 40 ft in 8 s. There is a direct proportionality between the two variables, x and t:

$\frac{10}{2}=\frac{40}{8}$

So, this means that at t = 0, the value of x is zero as well.

Also, we notice that the value of y increases by $\frac{75-30}{8-2}=7.5 ft/s$ (7.5 feet every second), so the initial value of y at t = 0 is:

$y(t=0)=30-7.5\cdot 2 =15 ft$

So, the initial position of the robot was (0,15) (15 feet above the southwest corner)

C)

The speed of the robot is given by

$v=\frac{d}{t}$

where d is the distance covered in the time interval t.

The distance covered is the one between the two points (10,30) and (40,75), so it is

$d=\sqrt{(40-10)^2+(75-30)^2}=54 ft$

While the time elapsed is

$t=8 sec-2 sec = 6 s$

Therefore the speed is

$v=\frac{54}{6}=9 ft/s$

D)

The equation for the line of the robot is:

$y=mx+q$

where m is the slope and q is the y-intercept.

The slope of the line is given by:

$m=\frac{75-30}{40-10}=1.5$

Which means that we can write an equation for the line as

$y=mx+q\\y=1.5x+q$

where q is the y-intercept. Substituting the point (10,30), we find the value of q:

$q=y-1.5x=30-1.5\cdot 10=15$

So, the equation of the line is

$y=1.5x+15$

E)

By prolonging the line above (40,75), we see that the line will hit the north wall. The point at which this happens is the intersection between the lines

$y=1.5x+15$

and the north wall, which has equation

$y=95$

By equating the two lines, we find:

$1.5x+15=95\\1.5x=80\\x=\frac{80}{15}=53.3 ft$

So the coordinates of impact are (53.3, 95).

F)

The distance covered between the time of impact and the initial moment is the distance between the two points, so:

$d=\sqrt{(53.5-0)^2+(95-15)^2}=95.7 ft$

From part B), we said that the y-coordinate of the robot increases by 15 feet/second.

We also know that the y-position at t = 0 is 15 feet.

This means that the y-position at time t is given by equation:

$y(t)=15+7.5t$

The time of impact is the time t for which

y = 95 ft

Substituting into the equation and solving for t, we find:

$95=15+7.5t\\7.5t=80\\t=10.7 s$

G)

The path followed by the robot is sketched in the second graph.

As the robot hits the north wall (at the point (53.3,95), as calculated previously), then it continues perpendicular to the wall, this means along a direction parallel to the y-axis until it hits the south wall.

As we can see from the sketch, the x-coordinate has not changed (53,3), while the y-coordinate is now zero: so, the robot hits the south wall at the point

(53.3, 0)

H)

The perimeter of the triangle is given by the sum of the length of the three sides.

- The length of 1st side was calculated in part F: $d_1 = 95.7 ft$

- The length of the 2nd side is equal to the width of the arena: $d_2=95 ft$

- The length of the 3rd side is the distance between the points (0,15) and (53.3,0):

$d_3=\sqrt{(0-53.3)^2+(15-0)^2}=55.4 ft$

So the perimeter is

$d=d_1+d_2+d_3=95.7+95+55.4=246.1 ft$

I)

The area of the triangle is given by:

$A=\frac{1}{2}bh$

where:

$b=53.5 ft$ is the base (the distance between the origin (0,0) and the point (53.3,0)

$h=95 ft$ is the height (the length of the 2nd side)

Therefore, the area is:

$A=\frac{1}{2}(53.5)(95)=2541.3 ft^2$

J)

The percentage of balls lying within the area of the triangle traced by the robot is proportional to the fraction of the area of the triangle with respect to the total area of the arena, so it is given by:

$p=\frac{A}{A'}\cdot 100$

where:

$A=2541.3 ft^2$ is the area of the triangle

$A'=90\cdot 95 =8550 ft^2$ is the total area of the arena

Therefore substituting, we find:

$p=\frac{2541.3}{8550}\cdot 100 =29.7\%$

4 0

3 years ago

Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all possibl

serg [7]

Answer:

P(X=i) where i = 1,2,3,4,5,6,7,8,9,10 = 1/2, 5/18, 5/36, 5/84, 5/252, 1/252, 0, 0, 0, 0.

Step-by-step explanation:

X denotes the highest ranking achieved by the woman.

When X=1, the top ranked person is a female.

When X=2, the first person is a male and the second ranked person is a female.

Similarly, when X=3, the first two ranked persons are male and the third one is a female.

When X=4, the first three persons are male and the fourth one is a female.

When X=5, the first four persons are males and the fifth person is a female.

When X=6, the first five people are males and the sixth person is a female. The rest of the four people are also females since there are only five men in a sample space.

The probability for X=7, 8, 9, 10 is zero because there are only five men who can achieve the first five positions and the last highest rank that can be achieved by a woman is 6.

To compute the probabilities, we will use the formula:

<u>No. of ways a female can be ranked X/Total number of ways to rank 10 people</u>

Note that the total number of ways of ranking 10 different people is 10P10 or 10!

For X=1, the first position can be taken by any of the 5 women. The possible ways of the first person being a woman is 5C1. The rest of the 9 people can take any of the ranks. They can be ordered in 9P9 ways.

So, P(X=1) = (5C1)(9P9)/(10P10) = (5 x 362880)/(3628800) = 1/2

For X=2, the first rank must be taken by a male. The number of ways to arrange the first person as a male out of the 5 men can be calculated by 5P1. The second position must be taken by a female and rest of the 8 positions can be taken by any of the 8 people in 8P8 ways.

So, P(X=2) = (5P1)(5C1)(8P8)/(10P10) = (5 x 5 x 40320)/(3628800) = 5/18

For X=3, first two people must be men and the number of ways to arrange 2 out of 5 males at the first two positions is 5P2. The third position is a female. The rest of the 7 people can be ordered in 7P7 ways.

P(X=3) = (5P2)(5C1)(7P7)/(10P10) = (20 x 5 x 5040)/(3628800) = 5/36

P(X=4) = (5P3)(5C1)(6P6)/(10P10) = (60 x 5 x 720)/(3628800) = 5/84

P(X=5) = (5P4)(5C1)(5P5)/(10P10) = (120 x 5 x 120)/(3628800) = 5/252

P(X=6) = (5P5)(5C1)(4P4)/(10P10) = (120 x 5 x 24)/(3628800) = 1/252

P(X=7) = 0

P(X=8) = 0

P(X=9) = 0

P(X=10) = 0

7 0

3 years ago