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laiz [17]
3 years ago
6

2. Jacqui is putting together sets of greeting cards for a school fundraiser. There are four different card options, two

Mathematics
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

32

Step-by-step explanation:

The problem statement states that there are four card option two colored envelopes options and four sticker design options and as the greeting card constitutes of one type of card.one colored envelopes and one sticker design then the number of ways Jacqui arrange the greeting card sets can be calculated using the counting principle that is n1*n2*n3. So, the number of ways Jacqui arrange the greeting card sets can be calculated using the counting principle=4*2*4=32 different ways.

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Find the Area of the figure below, composed of a rectangle and a semicircle. Round to the nearest tenth place.
kap26 [50]

Answer: 68.1

Step-by-step explanation:

The area of a semi-circle is 1/2 the area of a full circle. The area of a full circle is pi*r^2. Therefore, the area of a semi-circle is:

Area=\pi r^2/2

Area=\pi (3)^2/2=9\pi /2

The area of a rectangle is the length multiplied by the width:

Area=l*w

Area=6*9=54

The net area is the sum of the rectangle and the semi-circle:

Area=(9\pi /2)+(54)=68.1

8 0
2 years ago
Please help with imaginary numbers worksheet!
mihalych1998 [28]

Problem 5

<h3>Answer:   6i</h3>

------------------

Explanation:

We have these four identities

  • i^0 = 1
  • i^1 = i
  • i^2 = -1
  • i^3 = -i

Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.

We can think of it like this if you wanted

i^28 = (i^4)^7 = 1^7 = 1

Then the sqrt(-36) becomes 6i

So overall, we end up with the final answer of 6i

=============================================

Problem 6

<h3>Answer:  -3i</h3>

------------------

Explanation:

We'll use the ideas mentioned in problem 5

46/4 = 11 remainder 2

i^46 = i^2 = -1

sqrt(-9) = 3i

The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i

In other words, here is one way you could write out the steps

-i^{46}*-\sqrt{-9}\\\\-i^{2}*-3i\\\\i^{2}*3i\\\\-1*3i\\\\-3i\\\\

=============================================

Problem 7

<h3>Answer:   -1</h3>

------------------

Work Shown:

i^10 = i^2 because 10/4 = 2 remainder 2

i^19 = i^3 because 19/4 = 4 remainder 3

i^7 = i^3 because 7/4 = 1 remainder 3

Again, all we care about are the remainders.

i^{10}+i^{19} - i^{7}\\\\i^{2}+i^{3} - i^{3}\\\\i^{2}\\\\-1

=============================================

Problem 8

<h3>Answer:    -1 + i</h3>

------------------

Work Shown:

i^22*i^6 = i^(22+6) = i^28

Earlier in problem 5, we found that i^28 = i^0 = 1

So,

i^1 - \left(i^{22}*i^{6}\right)\\\\i^1 - i^{28}\\\\i^1 - i^{0}\\\\i - 1\\\\-1+i

8 0
2 years ago
Read 2 more answers
!!PLEASE HELP!! I WILL MARK U 5 STARS AND EVERYTHING
Sholpan [36]

Answer:

line 3 hope it helps :)

7 0
3 years ago
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Question 3(Multiple Choice Worth 2 points)
zysi [14]

The solution to the equation are 5 or -1 after applying the quadratic formula.

<h3>What is a quadratic equation?</h3>

Any equation of the form \rm ax^2+bx+c=0  where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

It is given that:

x² - 4x² - 5 = 0

The above equation is a quadratic equation:

Using the quadratic formula:

\rm  $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plug the values in the formula:

\rm  $x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}$

After solving:

x = 5 or -1

Thus, the solution to the equation is 5 or -1 after applying the quadratic formula.

Learn more about quadratic equations here:

brainly.com/question/17177510

#SPJ1

6 0
1 year ago
Find the circumference of the circle. Use calculator pl or 3.14. Round to the nearest hundredth, if necessary,
NISA [10]

Answer: 28.3

Step-by-step explanation:

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2 years ago
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