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Neporo4naja [7]
3 years ago
8

Solve the equation 46,207 x 83 = ___.

Mathematics
1 answer:
nadezda [96]3 years ago
5 0

Answer:

3,835,181

Step-by-step explanation:

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Rus_ich [418]

Answer:

The answer to 237 × 401 is <u>95037</u>

5 0
3 years ago
Read 2 more answers
Whats the answer to (-6p + 7) . -4?
KATRIN_1 [288]
-3 i think i used a calculator
7 0
3 years ago
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A swim teacher sells lesson packages. The best deal has the highest ratio of lessons to total cost.
tester [92]

Answer:

15 lessons for $ 80

Step-by-step explanation:

In case 1

Cost of 1 lesson = $10

In case 2

Cost of 1 lesson = 40/5 = $8

In case 3

Cost of 1 lesson = 80/10 = $8

In case 4

Cost of 1 lesson = 80/15 = $ 5.33

Hence,

The best deal is case 4 i.e 15 lessons for $ 80

3 0
3 years ago
Gerardo works at an ice cream shop each flavor of ice cream is stored in a right cylindrical tub that measures approximately 9 i
Leokris [45]

Answer:

Volume left = 404.32 cubic inches

Step-by-step explanation:

volume of a cylinder = \pir^{2}h

volume of ice cream in the cylindrical tub = \frac{22}{7} x 5^{2} x 9

                                           = \frac{22}{7} x 25 x 9

                                           = 707.1429

Volume of ice cream in the cylindrical tub is 707.14 cubic inches.

The ice cream scoop has a spherical volume, so that;

volume of a sphere = \frac{4}{3}\pir^{3}

r = \frac{2.5}{2}

 = \frac{5}{4} inches

volume of the scoop = \frac{4}{3} x \frac{22}{7} x (\frac{5}{4} )^{3}

                                  = \frac{4}{3} x \frac{22}{7} x \frac{125}{64}

                                  = 8.1845

volume of the scoop is 8.19 cubic inches.

If he scoops 37 scoops,

volume of the 37 scoops = 37 x 8.1845

                                          = 302.8265

volume of 37 scoops is 302.83 cubic inches.

Volume of ice cream left in the tub = 707.1429 - 302.8265

                                                           = 404.3164

Volume of ice cream left in the tub is 404.32 cubic inches.

7 0
3 years ago
Find the real or imaginary solutions of each equation by factoring
anyanavicka [17]
Sum of  2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so

x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4

the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
for x²-4x+16=0
x=\frac{-(-4)+/- \sqrt{(-4)^2-4(1)(16)} }{2(1)}
x=\frac{4+/- \sqrt{16-64} }{2}
x=\frac{4+/- \sqrt{-48} }{2}
x=\frac{4+/- (\sqrt{-1})(\sqrt{48}) }{2}
x=\frac{4+/- (i)(4\sqrt{3}) }{2}
x=2+/- 2i\sqrt{3}


the roots are
x=-4 and 2+2i√3 and 2-2i√3
8 0
3 years ago
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