$x^2-12x=-36\ \ \ |add\ 36\ to\ both\ sides\\\\x^2-12x+36=0\\\\x^2-2x\cdot6+6^2=0\ \ |use\ (a-b)^2=a^2-2ab+b^2\\\\(x-6)^2=0\iff x-6=0\ \ \ |add\ 6\ to\ both\ sides\\\\\boxed{x=6}$

6 0

3 years ago

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Find cscθ. pls answer

ExtremeBDS [4]

Answer:

The answer to your question is csc Ф = $\frac{-5}{4}$

Step-by-step explanation:

Process

1.- Determine the sign

We must determine csc Ф in the forth quadrangle, here csc is negative.

2.- Determine the hypotenuse

c² = a² + b²

c² = 3² + (-4)²

c² = 9 + 16

c² = 25

c = 5

3.- Determine csc Ф

csc Ф = $\frac{hypotenuse}{opposite side}$

csc Ф = $\frac{5}{-4}$ = $\frac{-5}{4}$

7 0

3 years ago

What is the value of x in the equation 1,331x3 − 216 = 0?

erma4kov [3.2K]

1331 x³ = 216
x³ = 216/1331
x = $\sqrt[3]{ \frac{216}{1331} } = \frac{6}{11}$
Thank you.

4 0

3 years ago

A Find tan a. r √5,-√7) ✓[? Enter

Virty [35]

Answer:

$tan(\alpha)=-\frac{\sqrt{35}}{5}$

Step-by-step explanation:

Tan can be defined as: $\frac{sin(\theta)}{cos(\theta)}$ as it simplifies to opposite/adjacent. If you know a bit about the unit circle, you'll know that the x-coordinate is going to be cos(theta) and the y-coordinate is going to be sin(theta). Since the sin(theta) is defined as opposite/hypotenuse, and the hypotenuse = 1, so sin(theta) is defined as the opposite side, which is the y-axis. Same thing goes for cos(theta), except the adjacent side is the x-axis.

Using this we can define tan

$sin(\alpha)=-\sqrt{7}\\cos(\alpha)=\sqrt{5}\\\\tan(\alpha)=-\frac{\sqrt{7}}{\sqrt{5}} * \frac{\sqrt{5}}{\sqrt{5}}\\tan(\alpha)=-\frac{\sqrt{7*5}}{5}\\tan(\alpha)=-\frac{\sqrt{35}}{5}\\$

8 0

2 years ago

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