A group of 32 players forms 4 volleyball teams. If there are 96 players, how many teams can be formed?

Answers

Answer 1
Answer:

There will be 12 teams can be formed if the group of 32 players forms 4 volleyball teams.

What is a fraction?

Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

We have:

A group of 32 players forms 4 volleyball teams.

Total number of players = 96×4

= 384

Number of teams = 384/32

= 12

Thus, there will be 12 teams can be formed if the group of 32 players forms 4 volleyball teams.

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Answer 2
Answer: 96/32=3
4*3=12


12 équipes volleyball if there are 96 players.


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One book costs one pound ninety five pence. How much does six books cost

Answers


All you have to do is times 195 by 6:

   195

x     6

-------------

1 1 7 0

Now make that back into £11.70

1.95*6=11.70   the answer is 11 pounds 70 pence

Y=−x−1 the point (0, −1)lies on the graph of the equation. the graph of the equation is the set of points that are solutions to the equation. the point (1, −2) lies on the graph of the equation. a coordinate pair on the graph of the equation is a solution to the equation. the equation y x=1 has the same graph.

Answers

For point (1, -2): -2 = -1 - 1 = -2. Therefore, point (1, -2) lies on the graph of the equation.

The graph of the equation is the set of points that are solutions to the equation.

A coordinate pair on the graph of the equation is a solution to the equation.

For the point (0, -1): -1 = -(0) - 1 = 0 - 1 = -1.
Therefore, The point ( 0, −1 ) lies on the graph of the equation.

A function is represented by the graph.Complete the statement by selecting from the drop-down menu.


The y-intercept of the function ​ y=−2x+8y​​ is the y-intercept of the function represented in the graph.
A graph with a line running through point (1,6) and point (5, -2)

Answers

Y = -2x + 8 The y intercept of the function is 8  

The equation that passes the point ( 1, 6 ) and ( 5, -2)
First solve the slope
M = ( 6 – (-2)) / ( 1 – 5 )
M = -2
At point ( 1,6) solve the y intercept b  
Y = mx + b
6 = (-2)(1) + b
b = 8
so the equation is   y = -2x + 8

Answer: (PIC)

Hope this helps- if so

PLEASE MARK BRAINIEST

Identify the number of solutions or zeros

2x^2+5x-9=0

Answers

There are going to be two solutions because the "biggest" x is x^2 but you can factor to find what they actually are.

Final answer:

The quadratic equation 2x^2+5x-9 = 0 has two solutions or zeros, which may be real or complex. They can be found using the quadratic formula: x = [-5 + sqrt(97)]/4 and x = [-5 - sqrt(97)]/4.

Explanation:

The equation given is 2x^2+5x-9=0, which is a quadratic equation. The solutions or zeros of a quadratic equation can be found using the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)]/2a. Here, a=2, b=5, and c=-9.

Let's substitute these values into the formula:

x = [-(5) ± sqrt((5)^2 - 4*2*(-9))]/2*2

x = [-5 ± sqrt(25 + 72)]/4

x = [-5 ± sqrt(97)]/4

Therefore, this equation has "two solutions" or zeros, which are x = [-5 + sqrt(97)]/4 and x = [-5 - sqrt(97)]/4.

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Find the third iterate, x3, of the function f(x) = 3x + 5 for an initial value of x0 = 1.

Answers

x3 = 4 because x0 = 1 
f(4) = 3.4 + 5 = 17
I think so

The rate of inflation is 3%. The cost of an item in future years can be found by iterating the function c(x)=1.03x. Find the cost of a $1500 refrigerator in three years if the rate of inflation remains constant.

Select one:

a. $1645.00

b. $1545.00

c. $1639.09

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Can someone help me in this trig question, please? thanks A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates 5pi/12

Answers

The exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

The position

Since the position of the carousel is (x, y) = (20cosθ, 20sinθ) and we need to find the position when θ = 5π/12 = 5π/12 × 180 = 75°

So, substituting the value of θ into the positions, we have

(20cos75°, 20sin75°)

The value of 20cos75°

20cos75° = 20cos(45 + 30)

Using the compound angle formula

cos(A + B) = cosAcosB - sinAsinB

With A = 45 and B = 30

cos(45 + 30) = cos45cos30 - sin45sin30

= 1/√2 × √3/2 - 1/√2 × 1/2

= 1/2√2(√3 - 1)

= 1/2√2(√3 - 1) × √2/√2

= √2(√3 - 1)/4

= (√6 - √2)/4

= (-√2 + √6)/4

So, 20cos75° = 20 × (-√2 + √6)/4

= 5 (-√2 + √6)

The value of 20sin75°

20sin75° = sin(45 + 30)

Using the compound angle formula

sin(A + B) = sinAcosB + cosAsinB

With A = 45 and B = 30

sin(45 + 30) = sin45cos30 + cos45sin30

= 1/√2 × √3/2 + 1/√2 × 1/2

= 1/2√2(√3 + 1)

= 1/2√2(√3 + 1) × √2/√2

= √2(√3 + 1)/4

= (√6 + √2)/4

= (√2 + √6)/4

So, 20sin75° = 20 × (√2 + √6)/4

= 5(√2 + √6)

Thus, (20cos75°, 20sin75°) = 5 (-√2 + √6), 5(√2 + √6).

So, the exact value of the position of the rider after the carousel rotates 5π/12 is 5 (-√2 + √6), 5(√2 + √6).

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\bf \textit{the position of the rider is clearly }20cos\left( (5\pi )/(12) \right)~~,~~20sin\left( (5\pi )/(12) \right)\n\n-------------------------------\n\n\cfrac{5}{12}\implies \cfrac{2+3}{12}\implies \cfrac{2}{12}+\cfrac{3}{12}\implies \cfrac{1}{6}+\cfrac{1}{4}\n\n\n\textit{therefore then }\qquad \cfrac{5\pi }{12}\implies \cfrac{1\pi }{6}+\cfrac{1\pi }{4}\implies \cfrac{\pi }{6}+\cfrac{\pi }{4}\n\n-------------------------------

\bf \textit{Sum and Difference Identities}\n\nsin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)\n\ncos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)\n\n-------------------------------\n\ncos\left( (\pi )/(6)+(\pi )/(4) \right)=cos\left( (\pi )/(6)\right)cos\left((\pi )/(4) \right)-sin\left( (\pi )/(6)\right)sin\left((\pi )/(4) \right)

\bf cos\left( (\pi )/(6)+(\pi )/(4) \right)=\cfrac{√(3)}{2}\cdot \cfrac{√(2)}{2}-\cfrac{1}{2}\cdot \cfrac{√(2)}{2}\implies \cfrac{√(6)}{4}-\cfrac{√(2)}{4}\implies \boxed{\cfrac{√(6)-√(2)}{4}}\n\n\nsin\left( (\pi )/(6)+(\pi )/(4) \right)=sin\left( (\pi )/(6)\right)cos\left( (\pi )/(4) \right)+cos\left( (\pi )/(6)\right)sin\left((\pi )/(4) \right)

\bf sin\left( (\pi )/(6)+(\pi )/(4) \right)=\cfrac{1}{2}\cdot \cfrac{√(2)}{2}+\cfrac{√(3)}{2}\cdot \cfrac{√(2)}{2}\implies \cfrac{√(2)}{4}+\cfrac{√(6)}{4}\implies \boxed{\cfrac{√(2)+√(6)}{4}}\n\n-------------------------------\n\n20\left( \cfrac{√(6)-√(2)}{4} \right)\implies 5(-√(2)+√(6))\n\n\n20\left( \cfrac{√(2)+√(6)}{4} \right)\implies 5(√(2)+√(6))