A=2(LW+LH+WH)

A=2((7/8)(1/3)+(7/8)(2/5)+(1/3)(2/5))

A=2(7/24+14/40+2/15)

A=14/24+28/40+4/15

A=7/14+7/10+4/15 210

A=(105+147+56)/210

A=308/210

A=(210+98)/210

A=1 98/210

A=1 7/15

8 0

2 years ago

Read 2 more answers

Here is a table of values y = f(x)

Rama09 [41]

A. Yes, the value of this function is 8 when x is 15.

B. False. I noted right away that the ordered pair (0,5) is not included, and some of the ordered pairs included do not represent the given function.

C. Yes, f(5) = 6. See (A) above.

D. False. The range of this function is the set of integers between 5 and 13 inclusive.

4 0

3 years ago

Andrew is about to leave for school. If he walks at a speed of 50 meters per minute, he will arrive 3 minutes after the bell rin

Mariana [72]

Answer:

The answer is: 13 minutes

Step-by-step explanation:

First Let us form equations with the statements in the two scenario

$time=\frac{distance}{speed}$

Let the time in which the bell rings be 'x'

1. If Andrew walks (50 meters/minute), he arrives 3 minutes after the bell rings. Therefore the time of arrival at this speed = (3 + x) minutes

$3 + x =\frac{distance}{50}\\distance = 50(3+x) - - - - - (1)$

2. If Andrew runs (80 meters/minute), he arrives 3 minutes before the bell rings. Therefore the time taken to travel the distance = (x - 3) minutes

$x - 3 = \frac{distance}{80} \\distance = 80(x-3) - - - - - (2)$

In both cases, the same distance is travelled, therefore, equation (1) = equation (2)

$50(3+x)=80(x-3)$

$150 +50x=80x-240\\$

Next, collecting like terms:

$150 + 240 = 80x - 50x\\390 = 30x\\30x = 390\\$

dividing both sides by 3:

x = 390÷30 = 13

∴ x = 13 minutes

7 0

2 years ago