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3 years ago

How do I work this out and explain how

Mathematics

1 answer:

Sav [38]3 years ago

4 0

80* 4/5

Cross out 80 and 5

You can divide by 5 for both of the numbers

80/5=16

5/5=1

16*4= 64

Answer is 64

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Carlos is mailing packages. Each small package costs him $2.40 to send. Each large package costs him $2.90. How much will it cos

aalyn [17]

Answer:

$11.10

Step-by-step explanation:

1 small package = $2.40

1 large package = $2.90

1 small package + 3 large packages.

2.40 + (3 × 2.90)

=2.40 + 8.70

=11.1

8 0

3 years ago

Read 2 more answers

Which number is rational , an integer , and a real number ?

vladimir2022 [97]

I believe the answer is C.

5 0

3 years ago

Geometry The picture below shows the question and answer choices for the questions.

ki77a [65]

Look at the table for the people that used Lithium.

There are 18 relapses, 6 No relapses with a total of 24 people.

The relative frequency for relapse, would be dividing the number of relapses by the total number of people.

This would be D. 18 / 24 = 75%

5 0

3 years ago

Which expression is a sum of cubes?

erik [133]

we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

Let's verify each case to determine the solution

case A) $-64x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-64=-4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case A) is not a sum of cubes

case B) $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

case C) $32x^{6} y^{12} +125x^{9} y^{3}$

we know that

$32=2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

case A) $64x^{6} y^{12} +125x^{9} y^{3}$

we know that

$64=4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

$4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}$

$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

the answer is

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

6 0

3 years ago

Read 2 more answers

-2x - y = -9 5x -2y = 18 Solve as a whole number

nexus9112 [7]

Answer:

x = 4, y = 1

(4, 1)

Step-by-step explanation:

-2x - y = -9

5x -2y = 18

-2x - y = -9

+2x +2x

-y = 2x - 9

/-1 /-1

y = -2x + 9

5x -2y = 18

5x - 2(-2x + 9) = 18

5x + 4x - 18 = 18

9x - 18 = 18

+ 18 + 18

9x = 36

/9 /9

x = 4

Now, we solve for y:

-2(4) - y = -9

-8 - y = -9

+ 8 +8

-y = -1

/-1 /-1

y = 1

(x, y) -> (4, 1)

Hope this helps!

6 0

2 years ago