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Varvara68 [4.7K]
3 years ago
9

If gasoline sells for $1.25 per gallon, how much would 3 3/5 gallons cost? express your answer as a decimal

Mathematics
2 answers:
diamong [38]3 years ago
8 0
3 3/5 gallons of gasoline would cost $4.50

This is because when you multiply 1.25 by 3 you get a product of 3.75.
Then, you multiple 1.25 with 3/5 and get a product of 0.75.
Add the two together, and you receive a sum of 4.50.
So, the cost for 3 3/5 gallon is $4.50.
blagie [28]3 years ago
5 0
I agree 100% good luck
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I know the answers but I don't know how to solve it
lutik1710 [3]
You did it right. 1.35 + 7.00 and you have the right answer. is that what you mean?
4 0
3 years ago
A fourth degree polynomial with real coefficients can have -4, 8i,+4i, and as its zeros. True or false? Justify your answer.
liraira [26]
You didn't give the fourth zero, but the answer is still false. If you have a root or an imaginary number as a zero, then its conjugate is also a zero. So if 8i is a zero, then -8i must also be a zero, and if 4i is a zero, then -4i must be a zero, with those zeros and -4, the number of zeroes exceeds the number of zeroes that a fourth degree polynomial can have.
4 0
3 years ago
Emma had 10 Balls in a bag. The balls were numbers 1 through 10. She wanted to test the probability of drawing a ball that is a
Goryan [66]

Answer:

  • 30%
  • Unlikely

Step-by-step explanation:

<u>The balls - multiples of 3 are:</u>

  • 3, 6 and 9

<u>The probability of a multiple of 3 is:</u>

  • 3 out of 10 or
  • P = 3/10 = 30%

30% probability is in the range 0%-50% so this event is unlikely

5 0
3 years ago
Read 2 more answers
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
What are the solutions to the quadratic equation x^2-16=0
Paha777 [63]

Answer:

x = ±4

Step-by-step explanation:

Hi there!

x^2-16=0

Move 16 to the other side

x^2=16

Take the square root of both sides

\sqrt{x^2}=\sqrt{16}\\x=\pm4

I hope this helps!

3 0
2 years ago
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