Round to the nearest ten 426,889

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

7 0

426,890
have a good day

liq [111]3 years ago

3 0

426,890
Hope this helps!!!!!!!!

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Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

4 years ago

Can someone pls help me <br> Solve for x <br> Please show ur work

Kitty [74]

Answer:

9=x

Step-by-step explanation:

80º=(8x+8)º

80=8x+8

Subtract 8 from both sides of the equation.

72=8x

Divide 8 from both sides of the equation.

9=x

Hope this helps :)

4 0

3 years ago

Mr. saslow has $32 for the class party. he spends $17 on the drinks and the cake. which of the inequalities represents the money

harkovskaia [24]

Answer:

I don't see any options for inequalities, so I wrote one based on the information.

Step-by-step explanation:

We'll assume that the drinks and cake ($17) are the only items Mr. Saslow needs to purchase and that the remainder may be used on whatever extras he wants. In this case, candy.

He starts with $32 and spends $17. The remainder may be spent on candy (all or none). This can be expressed as an inequality. Let C be the cost of the candy.

C $\leq$ ($32 - $17)