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eimsori [14]
3 years ago
6

The high school courtesy club bought red and black ink pens for Teacher Appreciation Day. The club bought 75 pens in all. Each b

lack pen cost $1.75, and each red pen cost $2.00. The club spent $137 altogether for the pens. How many black pens did the club buy?
13
23
48
52
Mathematics
1 answer:
dezoksy [38]3 years ago
8 0

Answer:

52

Step-by-step explanation:


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Examine the diagram and information to answer the question. Square ABCD has vertices at A(−2,1), B(2,7), C(8,3), and D(4,−3). Ho
BigorU [14]

Answer:

Option (1)

Step-by-step explanation:

Coordinates of the vertices are A(-2, 1), B(2, 7), C(8, 3) and D(4, -3)

Since ABCD is a square,

Perimeter of a square = 4 × (length of a side)

                                    = 4 × (AB)

Formula to calculate the distance between two points (x_1,y_1) and (x_2,y_2) is,

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Therefore, distance between two points A(-2, 1) and B(2, 7) will be,

AB = \sqrt{(2+2)^2+(7-1)^2}

AB = \sqrt{4^2+6^2}

AB = \sqrt{52}

AB = 2\sqrt{13}

Now area of  square ABCD = 4 × 2\sqrt{13}

                                              = 8\sqrt{13} unit

Therefore, option (1) will be the answer.

7 0
3 years ago
What property is illustrated by the statement 7+(4+4)=(7+4)+4
White raven [17]
Associative property of addition
<span> 7+(4+4)=(7+4)+4

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4 0
3 years ago
Item 11
Elena-2011 [213]

Answer:

54+2x=5x-3(2+6x)

54+2x=5x-6-18x

2x-5x+18x=-6-54

15x=-60

x=-60÷15

x=-4

3 0
4 years ago
Read 2 more answers
why do we need imaginary numbers?explain how can we expand (a+ib)^5. finally provide the expanded solution of (a+ib)^5.(write a
zheka24 [161]

Answer:

a. We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution.

b. (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

Step-by-step explanation:

a. Why do we need imaginary numbers?

We need imaginary numbers to be able to solve equations which have the square-root of a negative number as part of the solution. For example, the equation of the form x² + 2x + 1 = 0 has the solution (x - 1)(x + 1) = 0 , x = 1 twice. The equation x² + 1 = 0 has the solution x² = -1 ⇒ x = √-1. Since we cannot find the square-root of a negative number, the identity i = √-1 was developed to be the solution to the problem of solving quadratic equations which have the square-root of a negative number.

b. Expand (a + ib)⁵

(a + ib)⁵ =  (a + ib)(a + ib)⁴ = (a + ib)(a + ib)²(a + ib)²

(a + ib)² = (a + ib)(a + ib) = a² + 2iab + (ib)² = a² + 2iab - b²

(a + ib)²(a + ib)² = (a² + 2iab - b²)(a² + 2iab - b²)

= a⁴ + 2ia³b - a²b² + 2ia³b + (2iab)² - 2iab³ - a²b² - 2iab³ + b⁴

= a⁴ + 2ia³b - a²b² + 2ia³b - 4a²b² - 2iab³ - a²b² - 2iab³ + b⁴

collecting like terms, we have

= a⁴ + 2ia³b + 2ia³b - a²b² - 4a²b² - a²b² - 2iab³  - 2iab³ + b⁴

= a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴

(a + ib)(a + ib)⁴ = (a + ib)(a⁴ + 4ia³b - 6a²b² - 4iab³ + b⁴)

= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b + 4i²a³b² - 6ia²b³ - 4i²ab⁴ + ib⁵

= a⁵ + 4ia⁴b - 6a³b² - 4ia²b³ + ab⁴ + ia⁴b - 4a³b² - 6ia²b³ + 4ab⁴ + ib⁵

collecting like terms, we have

= a⁵ + 4ia⁴b + ia⁴b - 6a³b² - 4a³b² - 4ia²b³ - 6ia²b³ + ab⁴ + 4ab⁴ + ib⁵

= a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

So, (a + ib)⁵ = a⁵ + 5ia⁴b - 10a³b² - 10ia²b³ + 5ab⁴ + ib⁵

5 0
3 years ago
Write the fraction or decimal as a percent #1. 0.622 ,#2. 0.303 #3. 2.45
Pani-rosa [81]
So do you need it as a percent or a fraction?
8 0
3 years ago
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