In the graph below, what are all values of x

zimovet [89]

9514 1404 393

Answer:

88 square inches

Step-by-step explanation:

The figure can be divided into two congruent trapezoids, each with bases 7 and 4 inches, and height 8 inches. Then the total area is ...

A = 2(1/2)(b1 +b2)h

A = (7 +4)(8) = 88 . . . . square inches

4 0

3 years ago

During a scuba dive, Lainey descended to a point 20 feet below the ocean surface.She countined her descent at a rate of 20 feet

lorasvet [3.4K]

Answer:

$20 + 20x \leq 100$

Step-by-step explanation:

Let x be the number of minutes she can continue, since she's already 20 feet below the ocean surface and is diving at the rate of 20 ft/minute, her depth with respect to x would be

20 + 20x

But she doesn't want to reach more than 100 feet below the surface

$20 + 20x \leq 100$

4 0

3 years ago

Cancel the common factor of the numerator and the denominator and write specified expression

lisabon 2012 [21]

Step-by-step explanation:

Hello,

I hope you mean to cancel the common factor that exists in numerator and denominator,right.

so, Let's look for the common factor, 

here, the expression is, 

=4(x-2)/ (x+5)(x-2)

so, here we find the common factor is (x-2)

now, we have to cancel it. And after cancelling we get,

=4/(x+5)

Note:{ we cancel the common factor if the common factors are in multiply form.}

Hope it helps 

3 0

3 years ago

Suppose that there are 15 antennas in a store, of which 3 are defective. Assume all the defectives and all the functional antenn

ohaa [14]

Answer:

Step-by-step explanation:

Total number of antenna is 15

Defective antenna is 3

The functional antenna is 15-3=12.

Now, if no two defectives are to be consecutive, then the spaces between the functional antennas must each contain at most one defective antenna.

So,

We line up the 13 good ones, and see where the bad one will fits in

__G __ G __ G __ G __ G __G __ G __ G __ G __ G __ G __ G __G __

Each of the places where there's a line is an available spot for one (and no more than one!) bad antenna.

Then,

There are 14 spot available for the defective and there are 3 defective, so the arrange will be combinational arrangement

ⁿCr= n!/(n-r)!r!

The number of arrangement is

14C3=14!/(14-3)!3!

14C3=14×13×12×11!/11!×3×2

14C3=14×13×12/6

14C3=364ways

7 0

3 years ago

Some people think it is unlucky if the 13th day of a month falls on a Friday. Show that in every calendar year (non leap or leap

MakcuM [25]

We will set a variable, d, to represent the day of the week that January starts on. For instance, if it started on Monday, d + 1 would be Tuesday, d + 2 would be Wednesday, etc. up to d + 6 to represent the last day of the week (in our example, Sunday). The next week would start over at d, and the month would continue. For non-leap years:
If January starts on d, February will start 31 days later. Following our pattern above, this will put it at d + 3 (28 days would be back at d; 29 would be d+1, 30 would be d+2, and 31 is at d+3). In a non-leap year, February has 28 days, so March will start at d+3 also. April will start 31 days after that, so that puts us at d+3+3=d+6. May starts 30 days after that, so d+6+2=d+8. However, since we only have 7 days in the week, this is actually back to d+1. June starts 31 days after that, so d+1+3=d+4. July starts 30 days after that, so d+4+2=d+6. August starts 31 days after that, so d+6+3=d+9, but again, we only have 7 days in our week, so this is d+2. September starts 31 days after that, so d+2+3=d+5. October starts 30 days after that, so d+5+2=d+7, which is just d. November starts 31 days after that, so d+3. December starts 30 days after that, so d+5. Remember that each one of these expressions represents a day of the week. Going back through the list (in numerical order, and listing duplicates), we have d, d, d+1, d+2, d+3, d+3, d+3, d+4, d+5, d+5, d+6 and d+6. This means we have every day of the week covered, therefore there is a Friday the 13th at least once a year (if every day of the week can begin a month, then every day of the week can happy for any number in the month).
For leap years, every month after February would change, so we have (in the order of the months) d, d+3, d+4, d, d+2, d+5, d, d+3, d+6, d+1, d+4, and d+6. We still have every day of the week represented, so there is a Friday the 13th at least once. Additionally, none of the days of the week appear more than 3 times, so there is never a year with more than 3 Friday the 13ths.

5 0

3 years ago