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3 years ago

Please help me! I'm asking for help.

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

4 0

Answer:

Step-by-step explanation:

it's the 7.5*2*4.2 answer

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Which calculations could be used to solve this problem? There were 24 dogs and cats that were adopted from the animal shelter la

DENIUS [597]

Total number is 24.
Let the number of dogs be 3x and the number of cats be x.
3x + x = 24
4x = 24
x = 24/4
Number of dogs is : 24/4 × 3
Thus the answer is B.

Hope it helped!!

6 0

3 years ago

I do not know what this question is asking please help me

horsena [70]

So a median is the line segment joining one vertex to the midpoint of the opposite side. SO do that for all vertices, and see wehre they meet.

6 0

4 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

How many times can 36 go into 224

IRINA_888 [86]

$You\ must\ divide\ 224\ by\ 36.\\\\224|2\\112|2\\.\ 56|2\\.\ 28|2\\.\ 14|2\\.\ \ 7|7\\224=2\times2\times2\times2\times2\times7\\\\36|2\\18|2\\.\ 9|3\\.\ 3|3\\.\ 1\\\36=2\times2\times3\times3$

$224:36=\frac{\not2\times\not2\times2\times2\times2\times7}{\not2\times\not2\times3\times3}=\frac{56}{9}=\frac{54+2}{9}=\frac{54}{9}+\frac{2}{9}=\boxed{6\frac{2}{9}}$

8 0

3 years ago

Fatores 2m^4 - 18n^6

exis [7]

2m⁴ - 18n⁶
2(m⁴) - 2(9n⁶)
2(m⁴ - 9n⁶)
2(m⁴ - 3m²n³ + 3m²n³ - 9n⁶)
2[m²(m²) - m²(3n³) + 3n³(m²) - 3n³(3n³)]
2[m²(m² - 3n³) + 3n³(m² - 3n³)]
2(m² + 3n³)(m² - 3n³)

6 0

3 years ago