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BaLLatris [955]
3 years ago
15

The product of two negative integers is 36. The second integer is 5 more than the first. Find the integers.

Mathematics
1 answer:
Finger [1]3 years ago
8 0
-13 -23 because divide by two then add five and subtract 5
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Quadrilateral ABCD is an isosceles trapezoid. If || , AB = m + 6, CD = 3m + 2, BC = 3m, and AD = 7m – 16, solve for m.
pashok25 [27]

Answer:

m=2

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
X^3+3x^2y+3xy^2+y^3<br> ————————————<br> X^3+y^3
Hoochie [10]
To factor this fraction, you have be be aware of two special factoring formula:
a^3<span> + </span>b^3<span> = (</span>a<span> + </span>b)(a^2<span> – </span>ab<span> + </span>b^2<span>)

</span><span>(a+b)³ = a³ + 3a²b + 3ab² + b³

You can see the top part in this case is (x+y)^3, and the bottom (denominator) can be factor into (x+y)(x^2-xy+y^2)
we can cancel (x+y), so what we have left is (x+y)^2/(x^2-xy+y^2)
or (x^2+2xy+y^2)/(x^2-xy+y^2)
</span>
7 0
3 years ago
A survey of a group of students was conducted. The students were asked if they were taking a
sertanlavr [38]
<h3>Answer:  32</h3>

Explanation:

We add up the numbers found in the "math" circle

9+7+11+5 = 32

5 0
2 years ago
(x^2 - x^(1/2))/(1-x^(1/2))
Levart [38]
\frac { \left( { x }^{ 2 }-{ x }^{ \frac { 1 }{ 2 }  } \right)  }{ \left( 1-{ x }^{ \frac { 1 }{ 2 }  } \right)  }

\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x }  \right)  }{ \left( 1-\sqrt { x }  \right)  } \cdot 1

\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x }  \right)  }{ \left( 1-\sqrt { x }  \right)  } \cdot \frac { \left( 1+\sqrt { x }  \right)  }{ \left( 1+\sqrt { x }  \right)  }

\\ \\ =\frac { { x }^{ 2 }+{ x }^{ 2 }\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }

\\ \\ =\frac { -\sqrt { x } \left( 1-{ x }^{ 2 } \right) -x\left( 1-x \right)  }{ \left( 1-x \right)  }

\\ \\ =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right)  }{ \left( 1-x \right)  }

\\ \\ =\frac { \left( 1-x \right) \left\{ -\sqrt { x } \left( 1+x \right) -x \right\}  }{ \left( 1-x \right)  }

\\ \\ =-\sqrt { x } \left( 1+x \right) -x\\ \\ =-{ x }^{ \frac { 1 }{ 2 }  }\left( 1+{ x }^{ \frac { 2 }{ 2 }  } \right) -x

\\ \\ =-{ x }^{ \frac { 1 }{ 2 }  }-{ x }^{ \frac { 3 }{ 2 }  }-x\\ \\ =-\sqrt { x } -\sqrt { { x }^{ 3 } } -x
3 0
3 years ago
Read 2 more answers
Problem #7 PLEASE HELP ;(
mixer [17]

Answer:

Yes

6(x+0.4) is equivalent to 3(2x+0.8)

Step-by-step explanation:

Given in the questions two expressions

6(x + 0.4)

3(2x + 0.8)

We will apply distributive law

It is a law relating the operations of multiplication and addition, stated symbolically

<h3>a(b + c) = ab + ac</h3><h3 />

6(x + 0.4)

= 6(x) + 6(0.4)

= 6x + 2.4

3(2x + 0.8)

= 3(2x) + 3(0.8)

= 6x + 2.4

Since both equations when expanded have same answers, hence they are equivalent

4 0
2 years ago
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