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3 years ago

Solve for the zeros of the quadratic function f(x) = 9x2 + 6x + 1. Write the answer as a fraction.

2 answers:

4 0

We are asked to solve for the zeros of the given quadratic equation f(x) = 9x2 + 6x + 1. To solve this, first, we need to set the given equation into zero such as the succeeding solution is shown below:
0 = 9x2 + 6x +1
Perform factoring such as:
0 = (3x + 1) (3x + 1)
Solving for x1, we have:
3x + 1=0
3x = -1
x1 = -1/3
Soving for x2, we have:
3x +1=0
x = -1/3
Therefore, the zeroes of the given expression is both -1/3 for x1 and x2.

lutik1710 [3]3 years ago

4 0

the answer needed would be

-1/3

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3 years ago

One package of socks cost $7. How many packages can you but with $56?

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Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

Help quick! I'll mark your answer as the brainliest if your answer is correct!

Anna007 [38]

Answer:

They would never intersect because they are parallel lines. You can tell they are parallel lines because they have the same slope and they have a different y-intercept. However, this doesn't mean that there was no solutions, there were, thats how i was able to tell what their slopes and y-intercepts were. So if there is an answer that you can choose from that explains that they won't intersect because they are parallel then that is the correct answer. It may be C, but I would see if there is another option.

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2 years ago

Finding Derivatives Implicity In Exercise,Find dy/dx implicity.<br> x2e - x + 2y2 - xy = 0

Klio2033 [76]

Answer:

the question is incomplete, the complete question is

"Finding Derivatives Implicity In Exercise,Find dy/dx implicity . $x^{2}e^{-x}+2y^{2}-xy$ "

Answer : $\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

Step-by-step explanation:

From the expression $x^{2}e^{-x}+2y^{2}-xy$ " y is define as an implicit function of x, hence we differentiate each term of the equation with respect to x.

we arrive at

$\frac{d}{dx}(x^{2}e^{-x )+\frac{d}{dx} (2y^{2})-\frac{d}{dx}xy=0\\$

for the expression $\frac{d}{dx}(x^{2}e^{-x})$ we differentiate using the product rule, also since y^2 is a function of y which itself is a function of x, we have

$(2xe^{-x}-x^{2}e^{-x})+4y\frac{dy}{dx}-x\frac{dy}{dx} -y=0\\\\(2-x)xe^{-x}+(4y-x)\frac{dy}{dx}-y=0 \\$ .

if we make dy/dx subject of formula we arrive at

$(4y-x)\frac{dy}{dx}=y-(2-x)xe^{-x}\\\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

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3 years ago