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4 years ago

The first three steps of completing the square to solve the quadratic equation x^2+4x-6=0 are shown below.

1 answer:

5 0

Take the sqrt of both sides of <span>(x+2)^2=10:

x+2 = plus or minus sqrt(10)

Solve for x: x = -2 plus or minus sqrt(10)

Evaluate x: x = -2 + sqrt(10) AND x = -2 - sqrt(10) (answers)

</span>

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Find the area of an equilateral triangle using apothem length 6. PLEASE put in simplified radical form

marishachu [46]

Answer:

24√3

Step-by-step explanation:

8 0

3 years ago

1.4 has 2 significant digits as a result of which rule? a. All nonzero numbers are significant. b. Zeros between nonzero digits

marishachu [46]

Answer:

All non zero digits count in significant digits.

(a) is correct option.

Step-by-step explanation:

Given that,

1.4 has 2 significant digits.

We know that,

Significant digits :

All non zero digits count in significant digits.

All zero which is present in between two significant digits it is also significant.

Leading zero is not count as significant.

Trailing zero is count as significant.

We need to find 1.4 has 2 significant digits as a result of which rule

Using given rules

1.4 has 2 significant digits it is follows the rule of all non zero digits count in significant digits.

Hence, All non zero digits count in significant digits.

(a) is correct option.

6 0

3 years ago

Using the distributive property, write and equivalent expression for 7(x+2)

Marrrta [24]

Answer:

7x + 14

Step-by-step explanation:

3 0

3 years ago

May someone please help?

Vikki [24]

What do u need help on??

8 0

3 years ago

Find the derivative of <br> |x|/(x-1)

ki77a [65]

$\bf \cfrac{|x|}{x-1}\iff \cfrac{\sqrt{x^2}}{x-1}\iff \cfrac{(x^2)^{\frac{1}{2}}}{x-1}
\\\\\\
\textit{using the quotient rule}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{1}{2}(x^2)^{-\frac{1}{2}}\cdot 2x(x-1)-(x^2)^{\frac{1}{2}}\cdot 1}{(x-1)^2}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}}}{(x-1)^2}\implies \cfrac{dy}{dx}=\cfrac{x(x-1)-(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}(x-1)^2}
\\\\\\\cfrac{dy}{dx}=\cfrac{x(x-1)-|x|}{|x|(x-1)^2}$

4 0

3 years ago