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4 years ago

How do you factor x^2+11x+10=0?

Mathematics

1 answer:

xz_007 [3.2K]4 years ago

8 0

Answer:

x = -1 or x = -10

Step-by-step explanation:

x^2 + 11x + 10 = 0

x^2 + 10x + x + 10 = 0

x(x + 10) + 1(x + 10) = 0

(x + 1) (x + 10) = 0

x + 1 = 0 or x + 10 = 0

x = -1 or x = -10

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Lelechka [254]

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3 years ago

Graph the image of square CDEF after a reflection across the x-axis

Rina8888 [55]

Answer:

If you have a general point (x, y), and you reflect it across the x-axis, the coordinates of the new point will be:

(x,-y)

So we only change the sign of the y-component.

Now, if we do a reflection across the x-axis of a whole figure, then we apply the reflection to all the points that make the figure.

Then, we could just apply the reflection to the vertices of the square, then graph the new vertices, and then connect them, that is equivalent to graph the image of the square after the reflection.

The original vertices are:

C = (-3, 7)

D = (0, 7)

E = (0, 10)

F = (-3, 10)

Now we apply the reflection, remember that this only changes the sign of the y-component, then the new vertices are:

C' = (-3, -7)

D' = (0, -7)

E' = (0, - 10)

F' = (0, - 10)

Now we need to graph these points and connect them to get the reflected figure, the image can be seen below.

5 0

3 years ago

MATH HELP!!! 100PTS!!!

Lostsunrise [7]

Answer:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Step-by-step explanation:

<u>Given functions</u>:

$p(x)=2x-4$

$r(x)=\dfrac{6x-1}{9x+1}$

Solve for p(x) = r(x):

$\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}$

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

<u>Quadratic Formula</u>

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=18, \quad b=-40, \quad c=-3$

Substitute the values of a, b and c into the <u>quadratic formula</u> and solve for x:

$\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}$