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Citrus2011 [14]
2 years ago
14

Please help if you can

Mathematics
1 answer:
iogann1982 [59]2 years ago
8 0

-198.6 I believe because you take the square root of 704 because that is the product of -4 to the 5 power. then add nine. then multiply by 4. then subtract 15 they ta-da -198.6.

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paul paid $25.50 for 3 cups of hot chocolate and 4 cups of hot tea. the cost of each cup pf tea was 2.3 the cost of each cup of
Bumek [7]
How To Solve It:

$2.30 x 4 = $9.20

$25.50 - $9.20 = $16.30

$16.30 ÷ 3 = $5.43

Answer:

Each cup of hot chocolate costs $5.43

I hope this helps! :)
5 0
3 years ago
Graph the line y+3x=-3​
ivanzaharov [21]

Answer:  y = -3x-3 (Choose values for x)

Step-by-step explanation:

1st change the equation to y = mx+b form

In order to change the equation to y = mx+b form subtract the 3x to both sides.
The equation now becomes y = -3x-3

2nd we can either choose values for x such has (-2,-1,0,1,2) and plug in those values into our equation y = -3x-3
For example, if we plug in x= -2 we get y= -3(-2)-3 which becomes y =3. We then plot the point (-2,3) because -2 is our x coordinate and 3 is our y coordinate.
Or we can plug the equation into a calculator that will graph the equation for you.

3 0
1 year 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
A quadratic equation of the form 0 = ax2 + bx + c has one real number solution. Which could be the equation?
Fynjy0 [20]

Answer:

The equation showing this situation is  D=b^2-4ac=0

Step-by-step explanation:

Given : A quadratic equation of the form ax^2 + bx + c=0  has one real number solution.

To find : Which could be the equation?  

Solution :

A quadratic equation in form ax^2+bx+c=0 has a solution x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} called a quadratic formula  in which the roots are one real,two real or no real is determine by discriminant factor.

Discriminant is defined as to determine the number of roots in a quadratic equitation has following rules :

1) If D=b^2-4ac>0 there are two real roots.

2) If D=b^2-4ac=0 there are one real roots.

3) If D=b^2-4ac there are no real roots.

According to question,

A quadratic equation of the form ax^2 + bx + c=0  has one real number solution.

So, The equation showing this situation is  D=b^2-4ac=0

6 0
3 years ago
C+C+C+C+R=FF. When FxR= MF and CxC=MR. What is C equal to what is R equal to what is F equal to and what is M equal to?
Crazy boy [7]
I'm pretty sure R=M, hope that helps
8 0
3 years ago
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