Please Help!! Asap!!

A local club team is holding tryouts for six spots on the soccer team. Since there is not much time before a very important tournament they need to select the best players. Which method would be the best to ensure they select the "most talented" soccer players?


A. They watch the players during tryouts and then put them on a list in order from the most skilled to the least skilled and then select the top six from the list.


B. The first day of tryouts they select the first six that show up as they figure those are the most eager for a spot.


C. They place all of the names of the people trying out in a hat and then select six out of the hat. Those are the ones that are chosen.


D. They watch the players during tryouts and then put them on a list in order from the most skilled to the least skilled and then select the top three and the bottom three from the list.

Step-by-step explanation:

[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]

[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

We can write as :

27 = 3 × 3 × 3 = 3³

8 = 2 × 2 × 2 = 2³

243 = 3 × 3 × 3 × 3 × 3 = 3⁵

32 = 2 × 2 × 2 ×2 × 2 = 2⁵

[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now, we can write as :

(3³/2³) = (3/2)³

(3⁵/2⁵) = (3/2)⁵

[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now using law of exponent :

[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]

[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]

Answer:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

Answer:

3000

Step-by-step explanation:

Answer:

26/35

Step-by-step explanation:

1. First to divide the 3 1/2 by 2 3/5 you have to turn them both into improper fractions

First take 3 1/2. You have to multiply the whole number (3) by the denominator (2) and you would get 6. Then you would add then you add the product (6) to the numerator (1) and get 7.

You keep the denominator the same so the improper fraction is 7/2

Do the same thing to 2 3/5 and the improper fraction is 13/5

2. Now we can divide 13/5 by 7/2 using "keep, change, flip"

Keep: 13/5

Change: division to multiplcation

Flip: 7/2 to make 2/7

Your new equation is 13/5 × 2/7. Multiplcation is easy so you just have to multiply staight across: 13 × 2 and 5 × 7 giving you 26/35

If you divide 35 by 26 you will get 1.34 and a bunch of other numbers but I usually stop at two decimal places

hope this helps :)

9514 1404 393

Answer:

  7

Step-by-step explanation:

Of the 9 2-digit numbers ending in 6, only 2 are perfect squares: 16 and 36. The other 7 are not perfect squares.

Answer:

10

Step-by-step explanation:

3a+b - 4

Let a =3 and b = 5

3(3) +5 -4

Multiply

9+5-4

Add and subtract

14-4

10

Answer:

10

Step-by-step explanation:

= 3a + b - 4

= 3(3) + 5 - 4

= (3 . 3) + 5 - 4

= 9 + 5 - 4

= 9 + 1

= 10

Answer:

use graphs

Step-by-step explanation:

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

Answer:

a

Step-by-step explanation:

The transformation from Figure 1 to Figure 2 is:

The transformation of 5 units to the right and 3 units down.

Option C is the correct answer.

It is the movement of the shape in the left, right, up, and down directions.

The translated shape will have the same shape and shape.

There is a positive value when translated to the right and up.

There is a negative value when translated to the left and down.

We have,

A B C D E has points (-3, 5), (-2, 5), (-1, 4), (-2, 3), and (-5, 3).

A' B' C' D' E' has points (2, 2), (3, 2), (4, 1), (3, 0), and (0, 0).

Now,

A = (-3 + 5, 5 - 3) to A' = (2, 2)

B = (-2 + 5, 5 - 3) to B' = (3, 2)

C = (-1 + 5, 4 - 3) to C' = (4, 1)

D = (-2 + 5, 3 - 3) to A' = (3, 0)

E = (-5 + 5, 3 - 3) to E' = (0, 0)

We see that,

There is a translation of 5 units to the right and  3 units to the down.

Thus,

The transformation of 5 units to the right and 3 units down.

Learn more about translation here:

https://brainly.com/question/12463306

#SPJ1

Answer:

graph X only

Step-by-step explanation:

because with the rate of change it makes a straight line

Hello there! (:

The answer is 9.

3^4=3*3*3*3 (81)

3^2=9

81:9=9

So the answer is 9.

Hope it helps! If you have any question or query, feel free to ask! (:

~An excited gal

[tex]SparklingFlower[/tex]

Answer:

x=6

y=2

Step-by-step explanation:

-x+3y=0 (1)

x+3y=12 (2)

(1)+(2)<=> 6y=12

Answer: y=(1/5)m+3

Step-by-step explanation:

Start with the standard form of an equation of a straight line:  y=mx+b, whre m is the slope and b the y-intercept (the value of y when x=0).

y=mx+b

We know b (3), so:

y=mx + 3

To find the slope, m, we can use the one given point, (5,4):

y=mx + 3 for (5,4) would be:

4 = m*5+3

1 = 5m

m = (1/5)

y=(1/5)m+3

Answer:

I can't read that...........

Answer:

Chocolate for sure XD

Step-by-step explanation:

Hope this helps!! lol

Answer:

it is correct but -65 is not correct

Answer:

[tex]y=-\frac{1}{4}x\\[/tex]

Step-by-step explanation:

The slope is calculated by "up ÷ across".

= -1 ÷ 4

= [tex]-\frac{1}{4}[/tex]

The y-intercept is just 0 (because the line meets at the y axis at 0).

So, using [tex]y=mx+b[/tex] (where m = slope and b = y-intercept),

[tex]y=-\frac{1}{4}x+0[/tex]

but the '+0' is unnecessary so we just say [tex]y=-\frac{1}{4}x[/tex]

Answer:

There would be 12

Step-by-step explanation:

Answer:

[tex]\displaystyle m = 0[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Coordinate Planes

Step-by-step explanation:

*Note:

Rate of change is slope.

Step 1: Define

Identify.

Point (-1, -1)

Point (1, -1)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m.

Answer:

Step-by-step explanation:

Our even numbers from 1-9 are:

The largest possible number using the even numbers once is 8642.

Hoped this helped

Answer:

No inverse function:  (a), (b), (c)

Inverse function exists:  (d)

Step-by-step explanation:

The graph of h(x) = (x - 2)^2 + 3 is a parabola that opens upward and has vertex at (2, 3).  If the entire graph is drawn, and the horizontal line test then applied, h(x) would not have an inverse, because the horizontal line would intersect the  parabolic graph twice.

Note that if we restricted the domain to x ≥ 2, the resulting graph would pass the horizontal line test.  This would also be true for x ≥ 3, x ≥ 4, and so on.  Not so for (a) x < 1.  False for x > -5.  True for x < 3.  True for x > 4.

No inverse function:  (a), (b), (c)

Inverse function exists:  (d)

JK=JH

[tex]\\ \sf\longmapsto 10t=7t+15[/tex]

[tex]\\ \sf\longmapsto 10t-7t=15[/tex]

[tex]\\ \sf\longmapsto 3t=15[/tex]

[tex]\\ \sf\longmapsto t=15/3=5[/tex]

Answer:

A) No solution

Step-by-step explanation:

Given the systems of linear equations, y = 2x + 1 and y = 2x - 1:

Both equations in the system have the same slope, m = 2, thus forming parallel lines.  Since their lines are parallel from each other, then it means that their lines will never intersect.

Therefore, the given systems of linear equation is an inconsistent system that has no solution.

No I don't know your thing

Answer: 50

Step-by-step explanation:

exterior angles are supplement of interior angle therefore the 110 thing must be 70, also youre given 60 so 70+60=130

50 is missing from 180

Answer:

Make me as brain liest plz

The perimeter of triangle DEF is; 30 cm

We are given dimensions of Triangle ABC as;

Thus;

Perimeter of Triangle ABC = 9 + 6 + 7.5 = 22.5 cm

We are told that Triangle DEF is similar to Triangle ABC.

This means that the ratio of corresponding sides must be equal to the ratio of their perimeter.

Side EF corresponds to side BC. Thus;

Perimeter of Triangle DEF/22.5 = 8/6

Cross multiply to get;

Perimeter of Triangle DEF = (22.5 × 8)/6

Perimeter of Triangle DEF = 30 cm

Read more on similar triangles at; https://brainly.com/question/14285697