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3 years ago

The team practices 5.040

Mathematics

1 answer:

ikadub [295]3 years ago

8 0

Answer: the answer would be 12

Step-by-step explanation: it would be 12 because you are dividing the number uf players by the number of players on each team

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C&A Museum has three million visitors per year. Each visitor spends an averageof 1 hour in the museum. What is the average n

Alona [7]

Answer:

1250 visitors per day

Step-by-step explanation:

Number of visitors per year = 3,000,000

Number of days open per year = 300

Number of hours = 9a.m to 5pm = 8 hours

Number of visitors per hour :

Number of visitors per year / (number of days per year * number of hours per day)

3000000 / (300 * 8)

3000000 / 2400

= 1250 visitors per day

8 0

3 years ago

Evaluate the expression m=10 2m+10

Eddi Din [679]

The answer is 30
because (2)(10) + 10
20+10=30
the 10 came from the m!!

8 0

3 years ago

PLEASE HELP AND ANSWER!!!!! Which of the following reveals the minimum value for the equation 2x2 + 12x − 14 = 0?

NARA [144]

Answer:

The correct option is 3.

Step-by-step explanation:

The given equation is

$2x^2+12x-14=0$

It can be written as

$(2x^2+12x)-14=0$

Taking out the common factor form the parenthesis.

$2(x^2+6x)-14=0$

If an expression is defined as $x^2+bx$ then we add $(\frac{b}{2})^2$ to make it perfect square.

In the above equation b=6.

Add and subtract 3^2 in the parenthesis.

$2(x^2+6x+3^2-3^2)-14=0$

$2(x^2+6x+3^2)-2(3^2)-14=0$

$2(x+3)^2-18-14=0$

$2(x+3)^2-32=0$ .... (1)

Add 32 on both sides.

$2(x+3)^2=32$

The vertex from of a parabola is

$p(x)=a(x-h)^2+k$ .... (2)

If a>0, then k is minimum value at x=h.

From (1) and (2) in is clear that a=2, h=-3 and k=-32. It means the minimum value is -32 at x=-3.

The equation $2(x+3)^2=32$ reveals the minimum value for the given equation.

Therefore the correct option is 3.

5 0

3 years ago

Given f(x)=3x+12, find f(f^-1(0))

mamaluj [8]

Answer:

Step-by-step explanation:

METHOD 1.

We know:

$f\bigg(f^{-1}(x)\bigg)=x$

Therefore

$f\bigg(f^{-1}(0)\bigg)=0$

METHOD 2.

$\text{Find}\ f^{-1}(x)$

$f(x)=3x+12\to y=3x+12$

exchange x to y and vice versa:

$x=3y+12$

solve for y:

$3y+12=x$ <em> subtract 12 from both sides</em>

$3y=x-12$ <em>divide both sides by 3</em>

$y=\dfrac{x-12}{3}\to f^{-1}(x)=\dfrac{x-12}{3}$

$f\bigg(f^{-1}(x)\bigg)\to$ replace x in f(x) with the expression $\dfrac{x-12}{3}$

$f\bigg(f^{-1}(x)\bigg)=3\cdot\dfrac{x-12}{3}+12=x-12+12=x$

$f\bigg(f^{-1}(0)\bigg)\to$ put x = 0 to $f\bigg(f^{-1}(x)\bigg)=x$

$f\bigg(f^{-1}(0)\bigg)=0$

METHOD 3.

$\text{Find}\ f^{-1}(x)$

$f^{-1}(x)=\dfrac{x-12}{3}$

Calculate the value of <em>f</em> ⁻¹(x) for x = 0:

$f^{-1}(0)=\dfrac{0-12}{3}=\dfrac{-12}{3}=-4$

Calculate the value of <em>f(x)</em> for x = -4:

$f(4)=3(-4)+12=-12+12=0$

6 0

3 years ago

Need some help with this!

vredina [299]

Answer:

The answers is C

Step-by-step explanation:

I hope my answer hel you

8 0

3 years ago