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r-ruslan [8.4K]

2 years ago

10

Each basketball player runs 25 laps around the court during each practice. Which player has completed exactly of the distance fo

r this day of practice?

1 answer:

AVprozaik [17]2 years ago

8 0

Andy ran the complete distance I had this question before.

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Hey help me with this what's the answer of number 12

PtichkaEL [24]

What was the instructions given

6 0

2 years ago

Read 2 more answers

Help please thank you. will give brainliest. show all work and explain please.

alexandr402 [8]

divide the frequency of 6 by the total frequency to obtain a probability

$probability = \frac{9}{60} = \frac{3}{20}$

(also 15%)

8 0

3 years ago

Please help me!! Please explain and I’m pretty sure the answer needs to have < and > signs or >_ and <_ with “_” und

Julli [10]

Answer:

Jim has to work 48 hours to earn at least $300 per week.

Step-by-step explanation:

Jim makes 8.50 per hour

so in x hours he makes 8.5x

26% is deducted for taxes

so his take home pay is 8.5x(1 - .26) = 8.5x(.74) = 6.29x

To make > = $300 Jim must work enough hours so that

6.29x >= 300

x >= 17.49

4 0

3 years ago

Which equation is perpendicular to y = 4x - 6 and passes through point (8, 12).

sergeinik [125]

Answer:

The line is ;

4y = -x + 56

or

y = -x/4 + 14

Step-by-step explanation:

Generally, the equation of a straight line is;

y = mx + c

where m represents the slope

so for y = 4x - 6

The slope is 4

If two lines are perpendicular, the product of their slopes is -1

So for the second line, the slope will be -1/4

The equation in the point slope format will be ;

y-12 = -1/4(x-8)

y-12 = -x/4 + 2

Multiply through by 4

4y-48 = -x + 8

4y = -x + 8 + 48

4y = - x + 56

7 0

3 years ago

Consider f (x) = StartRoot x squared minus 1 EndRoot and g (x) = StartRoot x squared + 1 EndRoot. What value(s) of x would make

dalvyx [7]

Answer:

Any value of x

<em></em>

Step-by-step explanation:

Given

$f(x) = \sqrt{x^2 - 1}$

$g(x) = \sqrt{x^2 + 1}$

Required

What value of x is $f(g(x)) = g(f(x))$

Solving for f(g(x))

$f(x) = \sqrt{x^2 - 1}$

$f(g(x)) = \sqrt{(\sqrt{x^2 + 1})^2 - 1}$

Solve the inner square

$f(g(x)) = \sqrt{(x^2 + 1 - 1}$

$f(g(x)) = \sqrt{x^2 } }$

$f(g(x)) = x$

Solving g(f(x))

$g(x) = \sqrt{x^2 + 1}$

$g(f(x)) = \sqrt{(\sqrt{x^2 - 1})^2 + 1}$

$g(f(x)) = \sqrt{x^2 - 1 + 1}$

$g(f(x)) = \sqrt{x^2 }$

$g(f(x)) = x$

Equate f(g(x)) and g(f(x))

$f(g(x)) = g(f(x))$

$x = x$

<em>This implies that </em> $f(g(x)) = g(f(x))$ <em> at any value of x</em>

8 0

3 years ago