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xxTIMURxx [149]
3 years ago
14

Misty must pack 365 candles into 15 boxes. Each box must contain an equal number of candles. How many candles does she have left

over? (use long division to find the remainder)
A) 1
B) 3
C) 5
D) 7​
Mathematics
2 answers:
Korolek [52]3 years ago
5 0

Answer:

C

Step-by-step explanation:

10 can go into 365 10 times leaving 5 left over.

olganol [36]3 years ago
4 0

The answer is 5 because you can fit 365 into 15 24 times and then that would equal 360 and then you have 5 left over.

15x24=360             365

                              -  360

                                 ____

                                   5

You might be interested in
Divide using long division. Checky<br> (x3+ 3x2 + x - 9) = (x - 1)
Montano1993 [528]

Answer:

x squared plus 2x minus 1.

Step-by-step explanation:

x3+3x2+x-9 is in the division "house" and x-1 is on the outside. xsquare times x equals x cubed. and x square times negative 1 equals -x squsre. put x square on top of the roof. change the signs. the x cubed cancels out. and bring down 2x square from subtracting 3x square and -x square and bring down the x-9. Put 2x on top of the roof too and 2x times x equals 2x square and multiply by -1. U will have 2x square -2x square and -2x plus x which equals -1x cuz the 2x square cancels out. bring down the -9. Now i have -1x-9. Multiply by -1 and put -1 on the roof. -1x+1x cancels out. and -1 times -1 equals positive 1. So -9+1=8. U van do any further cause as soon as the last number is at the end of the roof and u cant go further, ur done dividing. so the answer is x square+2x-1/×-1 +8 as the remainder. hope i helped

6 0
3 years ago
How do you work this out in confused?
Leokris [45]
I think you either need to double 35 or use a protractor. Hope this is helpful!
4 0
3 years ago
Read 2 more answers
ASAP PLEASE HELP I BEEN WORKING ON IT FOR 5 HOURS
Luba_88 [7]
So in this case, the solution is where the graphs intersect. Put the equations in y-intercept form.
-4x+y=-6
y=4x-6

8x-2y=14
-2y=14-8x
y=4x-7

These lines have the same slope, which is 4. Therefore, they are parallel. Parallel lines never intersect, so there is NO SOLUTION.
7 0
3 years ago
Read 2 more answers
Enter an equation in point-slope form for the perpendicular bisector of the segment with endpoints M(−3, 7) and N(9, −3). The po
scZoUnD [109]

Answer:

5y - 6x = 53

Step-by-step explanation:

Given the segment with endpoints  M(−3, 7) and N(9, −3), let us find the slope first

m = y2-y1/x2-x1

m = -3-7/9-(-3)

m = -10/12

m = -5/6

Since the unknown line forms a perpendicular bisector, the slope of the unknown line will be:

m = -1/(-5/6)

m = 6/5

To get the intercept of the line, we will substitute m = 6/5 and any point on the line say (-3, 7) into the equation y = mx+c

7 = 6/5 (-3)+c

7 = -18/5 + c

c = 7 + 18/5

c = (35+18)/5

c = 53/5

Substitute m = 6/5 and c = 53/5

y = 6/5 x + 53/5

multiply through by 5

5y = 6x + 53

5y - 6x = 53

hence the point-slope equation of the perpendicular bisector is 5y - 6x = 53

6 0
3 years ago
If the volume of a box is 2x3 + 4x2 − 30xwhich of the dimensions are possible with the given x-value?
Kipish [7]

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is 2 x^{3} + 4 x^{2} -30x.

Step-by-step explanation:

The given is,

                        2 x^{3} + 4 x^{2} -30x................................(1)

Step:1

    Check for option A,

             x = 1, dimensions 8 by 9 by 1  

            From the equation (1),

                      Volume = 2 (1^{3}) + 4 (1^{2} )-30(1)

                                    =2+4-30 = -24...................(2)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                     = 72............................................(3)

            From equation (2) and (3)

                                -24 ≠ 72

            So, X=1; dimensions 8 by 9 by 1 is not possible.

   Check for option B,

             x = 1, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = 2 (1^{3}) + 4 (1^{2} )-30(1)

                                    =2+4-30 = -24...................(4)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30.........................................(5)

            From equation (4) and (5)

                                -24 ≠ 30

            So, X=1; dimensions 2 by 5 by 3 is not possible.

   Check for option C,

            x = 4, dimensions 2 by 5 by 3

            From the equation (1),

                      Volume = 2 (4^{3}) + 4 (4^{2} )-30(4)

                                    =2(64)+4(16)-30(4)

                                    = 128+64-120

                                    = 72.............................................(6)

            From the dimensions,

                      Volume = ( 2 × 5 × 3 )

                                     = 30............................................(7)

            From equation (6) and (7)

                               72 ≠ 30

            So, X=4; dimensions 2 by 5 by 3 is not possible.

    Check for option C,

            x = 4, dimensions 8 by 9 by 1

            From the equation (1),

                      Volume = 2 (4^{3}) + 4 (4^{2} )-30(4)

                                    =2(64)+4(16)-30(4)

                                    = 128+64-120

                                    = 72............................................(8)

            From the dimensions,

                      Volume = ( 8 × 9 × 1 )

                                    = 72............................................(9)

            From equation (8) and (9)

                               72 = 72

            So, X=4; dimensions 8 by 9 by 3 is possible.

Result:

           The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is 2 x^{3} + 4 x^{2} -30x.

         

4 0
3 years ago
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