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stepan [7]
3 years ago
8

Help me with this please show work please help!!!!!!!!!!!!!!!!!!!!!!...

Mathematics
2 answers:
Paul [167]3 years ago
8 0
The total cost to buy 2 bananas and 2 oranges is 1.70$
To find the cost of the bananas you would divide 1.50 by 6 and you arrive at 0.25 then each banana costs 25 cents, so if you want 2 it would cost you 50 cents.
To find the price of the oranges you would divide 3.00 dollars buy 5 and you arrive at 0.60 cents, so that means each orange costs .60 cents. If you want 2 oranges then you would add 60+60 and get 1.20$
Then you would add 50 cents and 1.20$ and arrive at 1.70$
Ulleksa [173]3 years ago
3 0
Correct me if im wrong but i think the answer is 4.5
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Find the range. 4.7 6.3 5.4 3.2 4.9 3.1 –3.1 9.5 –9.5
WINSTONCH [101]

Answer:

The range is 19

Step-by-step explanation:

First, you have to order the numbers from least to greatest:

-9.5, -3.1, 3.1, 3.2, 4.7, 4.9, 5.4, 6.3, 9.5

Then, you have to find the difference between the largest number and and smallest number. You do this by subtracting them:

9.5 - (-9.5)= 19

So, the range is 19

8 0
3 years ago
Write an equation in point-slope form for the line that has a slope of 4/5 and contains the point (−5, −3).
ankoles [38]

y = mx + b, where m = slope, and b = y-intercept.

Since you are not given the y-intercept, you have to solve for it through the equation, y = (4/5)x (which is the slope you are given) + b, and replace x and y with values given, which are (-5,-3)

y = (4/5)x + b

-3 = (4/5)(-5) + b

-3 = -4 + b

1 = b


Then replace b in the first equation to get the answer

y = (4/5)x + 1

5 0
3 years ago
3. The cost of a ticket twill be no more than $26.
goblinko [34]
Any number 1-26 should work. the expression would me x<$26.
7 0
3 years ago
Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam
kompoz [17]

For this case, we perform the conversions:

First roll:

1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x

x = \frac {3 * 100} {1}\\x = 300 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71

You can pack 71 boxes.

Second roll:

1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11

You can pack 11 boxes.

Thus, in total you can pack11 + 16 + 71 = 98 \ boxes

Answer:

98 boxes

4 0
3 years ago
Jody is lining the outer edge of her driveway with brick. The driveway is 4 yards long and 3 yards wide. In order for Jody to de
Triss [41]

A) Perimeter I’m 90% sure

I say that becuase if she is lining the OUTER EDGE of her driveway then she would need to know the outside og her driveway which would be perimeter.

8 0
3 years ago
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