Factoring x squared- 64= 0

2 answers:

Alex3 years ago

7 0

For this explanation, I'm going to write x squared as x^2.

This is whats called a "difference of squares" question. They appear whenever you have a squared x term (x^2), followed by a subtraction sign and a "perfect square" such as 4, 9, 25, 81, etc. It's actually very easy to factor! Determine the square root of the real number (in this case 64, whose square root is 8), then write two terms containing that number and a non-squared x. In one term separate them with a +, and the other with a -.

x^2 - 64 = 0
becomes
(x-8)(x+8) = 0

If you can remember your perfect squares, and can identify a question as a "difference of squares" question, they are really easy!

zvonat [6]3 years ago

6 0

Since −64-64 does not contain the variable to solve for, move it to the right side of the equation by adding 6464 to both sides.x2=64x2=64Take the square root of both sides of the equation to eliminate the exponent on the left side.x=±√64x=±64The complete solution is the result of both the positive and negative portions of the solution.
x=8,−8

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From a window 20 feet above the ground, the angle of elevation to the top of a building across

Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation: Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

$\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},$

and from the right-angled triangle WAB, we have'

$\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.$

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

8 0

3 years ago

Which of the following expressions is equivalent to (x + 5) ^2

Karo-lina-s [1.5K]

(X+5)^2 = x^2 +10x + 25

the expression that is equivalent to

To multiply it we apply FOIL method
Multiply x with x+5 first and then we multiply 5 with x+5

7 0

3 years ago

Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)

AlekseyPX

Answer:

The approximate solution to the system is $(-2.7, 2.1)$ .

Step-by-step explanation:

To solve the system of equations $\begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$ you must:

$\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$

$\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7$

$\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1$