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2 years ago

What is the value of the discriminant of the quadratic equation -2^3=-8x+8

Mathematics

1 answer:

rjkz [21]2 years ago

7 0

the value of the discriminant of the quadratic equation $-2x^2=-8x+8$ is $D =0$ .

Step-by-step explanation:

Here we need to find What is the value of the discriminant of the quadratic equation -2x^2=-8x+8 or , $-2x^2=-8x+8$ .

We know that for a quadratic equation , $f(x) = ax^2+bx+c$ , Discriminant is :

⇒ $D = b^2-4ac$

Now , let's frame equation $-2x^2=-8x+8$ in form of $f(x) = ax^2+bx+c$ :

⇒ $-2x^2=-8x+8$

⇒ $2x^2-8x+8=0$

⇒ $x^2-4x+4=0$

Comparing values we get that here , a=1 , b = -4 , c = 4 . Putting these values in $D = b^2-4ac$ :

⇒ $D = (-4)^2-4(1)(4)$

⇒ $D =16-16$

⇒ $D =0$

Therefore , the value of the discriminant of the quadratic equation $-2x^2=-8x+8$ is $D =0$ .

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How do you solve fractions and multiplication in algebraic, ex:1/3× -4

Setler79 [48]

Answer: x=12

Step-by-step explanation:

1/3 x -4

1/3x=4 <---- First add 4 to each side of the equation (moves the "-4")

1/3x(3)=4(3) <---- Muliply each side by "3" to get rid of the fraction

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3 years ago

For what values of the variables are the following expressions defined? 1. 5y+2 2. 18/y 3. 1/x+7 4. 2b/10−b Example: X>7

Serga [27]

Answer:

1. All real numbers

2. All real numbers except y = 0

3. All real numbers except x = -7

4. All real numbers except b = 10

Step-by-step explanation:

For any function to be defined at a particular value, it should not be approaching to a value $\infty$ or it should not give us the $\frac{0}{0}$ (zero by zero) form when the input is given to the function.

The value of function will depend on the denominator.

Now, let us consider the given functions one by one:

1. 5y+2

Here denominator is 1. So, it can not attain a value $\infty$ or $\frac{0}{0}$ (zero by zero) form

So, for all real numbers, the function is defined.

$2.\ \dfrac{18}{y}$

At y = 0, the value

$At\ y =0, \dfrac{18}{y} \rightarrow \infty$

So, the given function is defined for all real numbers except y = 0

$3.\ \dfrac{1}{x+7}$

Let us consider denominator:

x + 7 can be zero at a value x = -7

$At\ x =-7, \dfrac{1}{x+7} \rightarrow \infty$

So, the given function is defined for all real numbers except x = -7

$4.\ \dfrac{2b}{10-b}$

Let us consider denominator:

10-b can be zero at a value b = 10

$At\ b =10, \dfrac{2b}{10-b} \rightarrow \infty$

So, the given function is defined for all real numbers except b = 10

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3 years ago

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adoni [48]

Answer:

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Step-by-step explanation:

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3 years ago

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denpristay [2]

Answer: The answer is “10cm”

Step-by-step explanation: To get to this answer we first need to know the formula for find the hypotenuse which is a^2 + b^2 = c^2. A and B are the two sides in this case a and b are 8 cm and 6 cm. Then you plug in the values into the equation, it looks like this 8^2 + 6^ = c^2 once you solve you get 64 + 36 = c. When you add 64 and 36 you get 100 but their is one more step. You must find the square root of 100 which is 10. So your answer for the hypotenuse is “10cm”

Have a nice day!

4 0

2 years ago

What is the answer to 0=- 1/4(16)+b

Setler79 [48]

Answer:

Step-by-step explanation:

$0=-\dfrac{1}{4\!\!\!\!\diagup_1}\cdot16\!\!\!\!\!\diagup^4+b\\\\0=-4+b\qquad\text{add 4 to both sides}\\\\4=-4+4+b\\\\4=b\to b=4$

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2 years ago