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Eduardwww [97]
4 years ago
11

How many real solutions does 5x^2+3=-4 have

Mathematics
1 answer:
ruslelena [56]4 years ago
7 0
Hi,

Equation:

5 {x}^{2}  + 3 =  - 4

Since the left side is always >0, the equation has no solution.


Hope this helps.
r3t40
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What is h(x)=x^(2)+2 find x if h(x)=6 please help me
zysi [14]

Answer:

its both positive or negative 2

Step-by-step explanation:

Y=x^(2)+2

*subtract the 2 on both sides: 6=x^(2)+2

*square root both sides: 4=x^2

the square root of 4 is 2, so the answer would be positive 2.

we also know -2^2 is positive 4, which gives us

x=+2

x=-2

6 0
3 years ago
Is 2 DVDs / $36.75 equal to $18.375?
otez555 [7]

Just subtract those two numbers because then you will know if it is =2 that

3 0
4 years ago
Read 2 more answers
Use a property of equality to solve the equation <br> - 2a/5 = 13/15
vladimir2022 [97]

Answer:

a = -13/6

Step-by-step explanation:

Multiply both sides by the same number (15).  This clears out the fractions:

     -2a             13

15 -------- = 15 ----------

        5               15

Then we have -2a(3) = 13, or

                          -6a    = 13, or

                              a = -13/6

5 0
3 years ago
Circle O, with center (x, y), passes through the points A(0, 0), B(–3, 0), and C(1, 2). Find the coordinates of the center of th
balandron [24]
The answer:

the main formula of the circle's equation is 
(x-a)²+ (y-b)² = R²
where C(a, b) is the center of the circle
R is the radius

if a point A(x', y') passes through the circle, so the equation of the circle can be written as 
(x'-a)²+ (y'-b)² = R², and that is  a main formula.


<span>Circle O, with center (x, y), passes through the points A(0, 0), B(–3, 0), and C(1, 2), so we have exactly three equation:
</span>
(0-x)² + (0-y)² = R², circle O passes through A
x²+y²= R²
(-3 -x)² + (0-y)² = R², circle O passes through B
(-3 -x)² + (y)² = R²
(1-x)² + (2-y)² = R², circle O passes through A
(1-x)² + (2-y)² = R²

and we know that R= OA = OC= OB, 
OA=R= sqrt( (0-x)² + (0-y)² ) = OB = sqrt((-3 -x)² + (0-y)²), this implies

x²+y² = (-3 -x)² + (0-y)² , it implies x² = 9+ x² + 6x , and then -9/6=x, x= -3/2
and when OA = OC
x²+y² =(1-x)² + (2-y)²  so, x²+y² =1+x²-2x +4+y²-4y, therefore -5= -2x -4y
 -5= -2x -4y, when x = -3 /2   we obtain y = 2

the center is C(-3/2, 2)
7 0
4 years ago
X^3+3y-xy' = 0<br><br> dy/dx = xy^3-xy
Marat540 [252]

Answer:

dy/dx = xy^3-xy   =

y = ± √2 = ± 1.4142

x  =  0

y  =  0

Step-by-step explanation:

Step  1  : Simplify y/d

Step  2  :  See below

Step  3  :  See below

Pulling out like terms

Trying to factor as a Difference of Squares

Answer:  

y = ± √2 = ± 1.4142

x  =  0

y  =  0

<em><u>Hope this helps.</u></em>

6 0
4 years ago
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