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natta225 [31]
3 years ago
6

7 - 2x - x4

Mathematics
1 answer:
Vlad1618 [11]3 years ago
7 0

Answer:

b)  three terms

c)  -1

d)  7

e) 4

Step-by-step explanation:

put the expression in decreasing exponent order with the constant at the end

-x⁴ - 2x + 7

With this format it is easy to find the leading coefficient which is the coefficient of the term with the highest exponent; in this case you only see a negative sign but every variable, if no other coefficient is present, has an invisible '1'

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A complex number z_1z 1 ​ z, start subscript, 1, end subscript has a magnitude |z_1|=6∣z 1 ​ ∣=6vertical bar, z, start subscript
aliina [53]

Answer:

The answer is "\bold{9.19- 9.19\ i}"

Step-by-step explanation:

When the value of z_1 has the following properties:

\to |z_1| =13

\theta = 315^{\circ} \\\\ = 1.75 \pi\ \ \ \ \ \ \ \ \ \ \ \ \ \  \ _{where} \ \ (\pi = 180^{\circ})

Calculating the value of z_1 :

= 13 \times [ \cos(1.75 \pi ) + i \sin(1.75 \pi) ] \\\\=  13 \times[\cos(1.75 \pi - 2\pi ) + \ i \sin(1.75 \pi - 2\pi )]\\ \\= 13 \times [\cos(-0.25 \pi ) +\  i \sin(-0.25 \pi) ]\\\\= 13 \times [0.707106781 - 0.707106781]\\\\ =9.19 - 9.19\ i \\\\

3 0
3 years ago
The points (9, 9) and (5, 1) fall on a particular line. What is its equation in slope-intercept form?
Ilia_Sergeevich [38]
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D

Step-by-step explanation:

sohcahtoa

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3 0
3 years ago
2sin^2 2x=2<br><br> In the domain [0,2pi)<br><br> What can x equal ( in radians).
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Bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,


\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2}&#10;\\\\\\&#10;sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1}&#10;\\\\\\&#10;sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)

\bf \measuredangle 2x=sin^{-1}(\pm 1)\implies \measuredangle 2x=&#10;\begin{cases}&#10;\frac{\pi }{2}\\\\&#10;\frac{3\pi }{2}&#10;\end{cases}\\\\&#10;-------------------------------\\\\&#10;\measuredangle 2x=\cfrac{\pi }{2}\implies \measuredangle x=\cfrac{\pi }{4}\qquad \qquad \qquad \qquad \measuredangle 2x=\cfrac{3\pi }{2}\implies \measuredangle x=\cfrac{3\pi }{4}
3 0
3 years ago
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