I will give brainliest

Mathematics

1 answer:

DENIUS [597]3 years ago

4 0

Answer:

Candy | Number of People

Chocolate 10

Hard Candy 13

Chewy Candy 9

Step-by-step explanation:

We want to create a table that will show the data from the graph above.

We just have to read the frequencies of the corresponding candies of the bar graph.

The height of the bars tell us the frequencies.

Candy | Number of People

Chocolate 10

Hard Candy 13

Chewy Candy 9

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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
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

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3 years ago

Which statements are true for the given quadrilaterals? Choose all answers that are correct. A. All trapezoids are quadrilateral

svetlana [45]

Trapezoids always have 4 sides, so this is true. All rhombuses have parallel sides, so B is true also. A trapezoid has two sides that are not parallel, so C is not true. (remember, you only have to find one instance where your statement is false for it to be false) And for D, all rhombuses are kite-shaped so D is not true. A and B are true. Hope this helps. :)

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3 years ago

Read 2 more answers

a store mixes red fescue worth $12 per pound and chewings fescue worth $16 per pound. The mixture is to sell for $15 per pound.

nlexa [21]

Answer is: 414 pounds of red fescue and 138 pounds of chewings fescue is needed to produce 552 pounds of mixture.

Explanation:
x - how much of red fescue is needed for one pound of mixture
y - how much of chewings fescue is needed for one pound of mixturex*12+ y*16 = 15x + y = 1
so, x=1-y
putting it into first line:(1-y)*12 + y*16 = 15
12 - 12*y + 16*y = 15
4*y = 3
y=3/4
so, x= 1-3/4; x=1/4So to produce 552 pounds of mixture we need..To produce 1 pound of mixture we need: 3/4 pound of red fescue and 1/4 pound of chewing fescue so to produce 552 pounds of miture we need:
552*3/4 - red fescue, which is 414 pounds needed
and
552/4 - chewings fescue, which is 138 pounds needed