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3 years ago

PLEASE HELP!!! I've been really struggling with these. Can someone assist with this?

Mathematics

1 answer:

Archy [21]3 years ago

4 0

When dealing with little odds you must know one part and the whole part. For number 1 a fraction , 7 is a part of a 11 which is the whole. For 2 a percent, 6% part and 100% whole, here you had to know that for 6% to exist there had to be a 100% . For 3 a ratio, the first number is part (2) and the second is whole (7). Look at pic for all work.

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How do you find the radius of a circle using the circumference

Elanso [62]

Answer:

r=C/2π

Step-by-step explanation:

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3 years ago

Please explain how to get answer in sq units of parallelogram ABCD

Lostsunrise [7]

Hey there :)

Assuming you are asking how to find the area of parallelogram ABCD:

We know the area formula of a parallelogram:
Area = base x height

Then apply the formula to the given.

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3 years ago

If 3m - n = 11, then m =

irina [24]

Answer:

n= -2 and m=3

Step-by-step explanation:

3 x 3= 9

11- 9 = 2

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3 years ago

The judiciary committee at a college is made up of three faculty members and four students. If ten faculty members and 25 studen

Andre45 [30]

Answer:

1,518,000 committees.

Step-by-step explanation:

We have been given that the judiciary committee at a college is made up of three faculty members and four students. Ten faculty members and 25 students have been nominated for the committee.

We will use combinations for solve our given problem.

$^nC_r=\frac{n!}{r!(n-r)!}$ , where,

n = Number of total items,

r = Items being chosen at a time.

$^{10}C_3\cdot ^{25}C_4=\frac{10!}{3!(10-3)!}\cdot \frac{25!}{4!(25-4)!}$

$^{10}C_3\cdot ^{25}C_4=\frac{10!}{3!*7!}\cdot \frac{25!}{4!*21!}$

$^{10}C_3\cdot ^{25}C_4=\frac{10*9*8*7!}{3*2*1*7!}\cdot \frac{25*24*23*22*21!}{4*3*2*1*21!}$

$^{10}C_3\cdot ^{25}C_4=10*3*4\cdot 25*23*22$

$^{10}C_3\cdot ^{25}C_4=120\cdot 12650$

$^{10}C_3\cdot ^{25}C_4=1,518,000$

Therefore, 1,518,000 judiciary committees could be formed at this point.

7 0

3 years ago

What is the correct justification for the indicated steps?

yanalaym [24]

The given proof of De Moivre's theorem is related to the operations of

complex numbers.

<h3>The Correct Responses;</h3>

Step A: Laws of indices

Step C: Expanding and collecting like terms

Step D: Trigonometric formula for the cosine and sine of the sum of two numbers

<h3>Reasons that make the above selection correct;</h3>

The given proof is presented as follows;

$\mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1}}$

Step A: By laws of indices, we have;

$\left[cos(\theta) + i \cdot sin(\theta) \right]^{k + 1} = \mathbf{\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}$

$\left[cos(\theta) + i \cdot sin(\theta) \right]^{k} \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] = \mathbf{\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right]}$

Step B: By expanding, we have;

$\left[cos(k \cdot \theta) + i \cdot sin(k \cdot \theta) \right] \cdot \left[cos(\theta) + i \cdot sin(\theta) \right] = cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right]$

Step D: From trigonometric addition formula, we have;

cos(A + B) = cos(A)·cos(B) - sin(A)·sin(B)

sin(A + B) = sin(A)·cos(B) + sin(B)·cos(A)

Therefore;

$cos(k \cdot \theta) \cdot cos(\theta) - sin(k \cdot \theta) \cdot sin(\theta) + i \cdot \left [sin(k \cdot \theta) \cdot cos(\theta) + cos(k \cdot \theta) \cdot sin(\theta) \right] = \mathbf{ cos(k \cdot \theta + \theta) + i \cdot sin(k \cdot \theta + \theta)}$

Learn more about complex numbers here:

brainly.com/question/11000934

4 0

2 years ago

Read 2 more answers