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3 years ago

I need help what is the correct answer

Mathematics

2 answers:

leonid [27]3 years ago

8 0

42/52 is what i calculated

kondaur [170]3 years ago

8 0

14 out of 47

or 30%

(.297)

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(100 Points) Please answer (For) This question guys what is (Free) 1+1

Arada [10]

Answer:

Step-by-step explanation:

1 + 1 = 2

4 0

3 years ago

Read 2 more answers

X-12y=-210 and x-6=90

m_a_m_a [10]

Answer:

x = -30

y = -20

Step-by-step explanation:

x - 12y = 210..... (I)

x - 6y = 90..... (II)

Subtract equation (II) from (I)

- 12y - (-6y) = 210 - 90

-12y + 6y = 120

-6y = 120

Divide through by -6y

Therefore, y = -20

Substitute y = -20 in equation (II)

x -6 (-20) = 90

x + 120 = 90

x = 90 - 120

x = -30

Hope that helped

7 0

3 years ago

A quadratic pattern has a second term equal to 1,a third term equal to -6 and a fifth term to -14 . calculate the second differe

PolarNik [594]

Answer:hi f

Step-by-step explanation:

Kligcddfcgg

8 0

3 years ago

PLLSS HELPPP itss due nowww

allsm [11]

Answer:

libk is given check it out

Step-by-step explanation:

https://www.slader.com/discussion/question/beginarray-l-triangle-t-a-m-has-vertices-t-05-a-41-and-m-36-find-the-gide-reflection-image-of-delta/

7 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