The point (-6, 4) is reflected over the x-axis. What is the new point?

Mathematics

1 answer:

defon2 years ago

7 0

Answer:

p (6, -4)-------------->p'(6,4)

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Which has a value closest to 9?

almond37 [142]

Answer:

Step-by-step explanation:

4 0

3 years ago

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Consider the following "generic rectangle" where the area of each region is given. Write equivalent

marta [7]

Answer:

sum: 3x² -4x -4
product: (x -2)(3x +2)

Step-by-step explanation:

The areas of four regions are given. We can simply add them to find the sum. To express them as a product, we need to look at common factors.

The total of the given area expressions is ...

3x² +2x -6x -4 = 3x² -4x -4 . . . . sum

<h3>Product</h3>

Extending the table to show common factors of each row and column, we have ...

$\begin{array}{c|c|c|}&3x&2\\\cline{1-3}x&3x^2&2x\\\cline{1-3}-2&-6x&-4\\\cline{1-3}\end{array}$

Since each cell of the table is the product of the corresponding common factors, we can write the area as the product ...

(x -2)(3x +2) . . . . product

8 0

1 year ago

If x^2+1/x^2=3 find the value of x^2/(x^2+1)^2<br> Express answer as a common fraction.<br> Thanks!!

timama [110]

$x^2+\dfrac1{x^2}=3\implies x^4+1=3x^2\implies x^4-3x^2+1=0$

By the quadratic formula,

$x^2=\dfrac{3\pm\sqrt5}2\implies x^2+1=\dfrac{5\pm\sqrt5}2$

Then

$(x^2+1)^2=\dfrac{25\pm10\sqrt5+5}4=\dfrac{15\pm5\sqrt5}2$

$\implies\dfrac{x^2}{(x^2+1)^2}=\dfrac{\frac{3\pm\sqrt5}2}{\frac{15\pm5\sqrt5}2}=\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}$

Multiply numerator and denominator by the denominator's conjugate:

$\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}\cdot\dfrac{15\mp5\sqrt5}{15\mp5\sqrt5}=\dfrac{45\pm15\sqrt5\mp15\sqrt5-25}{15^2-(5\sqrt5)^2}=\dfrac{20}{100}=\dfrac15$