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Shkiper50 [21]
3 years ago
9

There are 20 boys and 35 girls that want to participate in the school play. If each set must have the same ratio of boys to girl

s, what is the greatest number of sets that can be in the play. Find how many boys and girls each set will have
Mathematics
1 answer:
faust18 [17]3 years ago
4 0
The greatest number of sets would be 20 with one girl and one boy in each.
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Max and Sharifa are both saving to buy the same model of DVD player, which costs $99, including tax. Max already has $31.00 and
gogolik [260]

Answer:

Sharifa can pay for the DVD player first.

Step-by-step explanation:

<u>Max:</u>

99 - 31 = $68 remaining

68 / 8.5 = <em>8 weeks</em>

<u>Sharifa:</u>

99 - 25.5 = $73.50 remaining

73.5 / 10.5 = <em>7 weeks</em>

7 0
3 years ago
PLEASE HELP ILL GIVE 49 POINTS! THANKS!<br><br><br> Show work
NemiM [27]

Answer:

y = -2x+11

Step-by-step explanation:

If we have a slope and a line, we can use the slope intercept form of a line

y-y1 = m(x-x1)

where (x1,y1) is the point and m is the slope.


Substitute in what we know

y-5 = -2 (x-3)

We want it in slope intercept form which is y = mx+b

Distribute the -2

y-5 = -2x+6

Add 5 to each side

y-5+5 = -2x +11

3 0
3 years ago
Read 2 more answers
A rectangular shoebox has a volume of 728 cubic inches.the base of the shoebox measures 8 inches by 6.5 inches.how long is the s
Flauer [41]

Answer:the answer is 14

Step-by-step explanation: first you want to multiply

8*6.5

=52

then, since you want to get x being what it takes to get to 728 you divide

728/52

14

hope this helps

Goog luck!

5 0
3 years ago
I would like to create a rectangular vegetable patch. The fencing for the east and west sides costs $4 per foot, and the fencing
Anestetic [448]
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y.  the dimensions you want are therefore x and y.

The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y.  We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.



The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128.  Let's reduce this by dividing all three terms by 4:  2x + y = 32.

Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32.  The formula for area is A = L * W.  Solving 2x + y = 32 for y, we get y = -2x + 32.

We can now eliminate y.  The area of the patch is (x)(-2x+32) = A.  We want to maximize A.

If you're in algebra, find the x-coordinate of the vertex of this quadratic equation.  Remember the formula x = -b/(2a)?  Once you have calculated this x, subst. your value into the formula for y:  y= -2x + 32.

Now multiply together your x and y values to obtain the max area of the patch.


If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero.  This approach should give you the same x value as before; the corresponding y value will be the same;  y=-2x+32.

Multiply x and y together.  That'll give you the maximum possible area of the garden patch.
5 0
3 years ago
Help please answer for this (16/81) - 3/4
kozerog [31]

Answer:

Here is the answer

Step-by-step explanation:

-0.552469136

Btw brainliest me plssssssss

6 0
3 years ago
Read 2 more answers
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