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Zarrin [17]
2 years ago
5

What is the solution to this equation ?

Mathematics
1 answer:
iren2701 [21]2 years ago
6 0

Answer:

x = -20/9

Step-by-step explanation:

First we need to distribute this equation 4(4x+1) = 3x-4(6-x)+8. After distributing we get 16x+4 = 3x-24+4x+8. Next we add like terms to get 16x+4 = 7x-16. Now subtract 7x on both sides to get 9x+4 = -16. Now subtract -4 on both sides to get 9x = -20. Now divide by 9 on both sides to get x= -20/9.

Hope that helps! :)

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Nine less than three times a number is greater than forty
pentagon [3]
Start with an equation.
Nine less... so -9
3 times a number = 3x
Greater than 40 is >40

So put together is 3x-9>40
When solving I imaging the greater than sign as an equal sign. Less confusing.
3x-9=40
3x=49
x>49/3
7 0
3 years ago
25/75 = x/15 what is x
liq [111]

Answer:

x=5

Step-by-step explanation:

8 0
3 years ago
One more... and ill be out of yalls hair.
Zarrin [17]
Awnser:69
I had this question a few minutes ago, that’s funny
7 0
2 years ago
If a quadratic equation has 4-i as a solution, what must the other solution be?
Luden [163]

Answer:

4+i

Step-by-step explanation:

A complex number usually took the form a+bi where a and b are real numbers and 'i' represents an imaginary number. For a quadratic equation, the complex roots for the root of a quadratic equation took the form known as complex conjugates. The complex conjugates are formed by changing the sign of the imaginary part.

SO, if a quadratic equation has 4-i as a solution, the other solution must be 4+i.

6 0
2 years ago
Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans
zloy xaker [14]
Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.



Part B:

The area of the square is given by

Area=(x+4)^2=x^2+8x+16

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

x^2+8x+16=6x+8 \\  \\ \Rightarrow x^2+8x-6x+16-8=0 \\  \\ \Rightarrow x^2+2x+8=0 \\  \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2}  \\  \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2}  \\  \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
7 0
3 years ago
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