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3 years ago

14

What is a factor of 144-49x^2

1 answer:

Alenkinab [10]3 years ago

3 0

(12+7x)(12-7x)
Because the square root of 144 is 12 and for 49 is 7

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At what point do the lines y = 6x – 18 and<br> y=8x – 32 intersect?

anzhelika [568]

Answer:

(7, 24)

Step-by-step explanation:

Solve the system:

6x - 18 = 8x - 32

2x = 14

x = 7

y = 6(7) - 18

y = 24

4 0

3 years ago

Dahasolnce [82]

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$ , b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

$\sqrt{3}\cdot \csc x - 2 = 0$

$\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0$

$\sqrt{3} - 2\cdot \sin x = 0$

$\sin x = \frac{\sqrt{3}}{2}$

$x = \sin^{-1} \frac{\sqrt{3}}{2}$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

b) The equation must be simplified first:

$\cos x + 1 = - \cos x$

$2\cdot \cos x = -1$

$\cos x = -\frac{1}{2}$

$x = \cos^{-1} \left(-\frac{1}{2} \right)$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

7 0

3 years ago

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Arisa [49]

Answer:

Part A - $2\frac{1}{2} + b = 4\frac{2}{3}$

Part B - $2\frac{1}{6}$

Step-by-step explanation:

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