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Setler [38]
3 years ago
10

The value of G for a one-piece 30° x 60° angle iron bracket is 1'/4 inches, and the width of the angle iron is 2 inches. What is

the thickness of the metal, in inches? A. 31/4 inches B. 3/4 inch C. 21/2 inches D. 1/4 inch
Mathematics
1 answer:
Alborosie3 years ago
8 0

Answerb

Step-by-step explanation:

yeal

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Two ships leave a harbor together traveling
Andrew [12]

Answer:

use cosine law

x^2 = 543^2 + 543^2 - 2(543)(543cos 128°

solve for x

Step-by-step explanation:

5 0
2 years ago
10,000 candy canes was divided into 50 boxes. How many candy canes were in each box?​
kumpel [21]

Answer:

200

Step-by-step explanation:

5 0
2 years ago
Julio's father is 4 times as old as Julio. The sum of their ages is no less than 55.
seropon [69]
Call (F) the age of the father and (J) the age of Julio

The F & J are related in this way: F=4J

 Now you have a restriction in the form of inequality: The sum of both ages has to be greater or equal than 55.

Algebraically that is: F + J ≥ 55

You can substitute F with 4J to find the solution for J: 

4J + J ≥ 55

5J ≥ 55

Now divide both sides by 5

5J/5 ≥ 55/5

J ≥ 11 

That Imposes a lower boundary for the value of J of 11, meaning that the youngest age of Julio can be 11
8 0
3 years ago
A) How many ways can 2 integers from 1,2,...,100 be selected
Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

7 0
3 years ago
PLEASE HELP I WILL MARK
lys-0071 [83]

Answer: 37 pieces of candy

Step-by-step explanation:

24-8.50 is 15.50 she wants to save 6.25, so, 15.50-6.25 is 9.25. 9.25 divided by the cost of each candy (0.25) is 37

6 0
3 years ago
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