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zvonat [6]
3 years ago
8

Give an example of an open equation​

Mathematics
1 answer:
rusak2 [61]3 years ago
4 0

the example of an open equation is :x+9=16

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PLZ HELP THIS IS DUE TODAY AND NO LINKS PLZ.
anastassius [24]

Answer:

D is the best answer to the question

8 0
2 years ago
Thomas creates the table shown to represent the function ƒ(x) = x2 + 3x - 10. determine the zeros of the equation.
Eddi Din [679]
Answer: A

Explanation: The zeroes of f(x) are the value/s of x such that f(x) = 0. So, we need to find the values of x in the equation f(x) = 0.

Note that

f(x) = 0
⇔ x² + 3x - 10 = 0

By factoring into binomials, x² + 3x - 10 = (x + 5)(x - 2). Thus, 

x² + 3x - 10 = 0
⇔ (x + 5)(x - 2) = 0
⇔ x + 5 = 0 or x - 2 = 0
⇔ x = -5 or x = 2

Hence, the zeroes of x are  -5 and 2. 
4 0
3 years ago
Find the area of the following parallelograms
Kay [80]

Answer:

Area of parallelogram: 1/2×h×sum of base lengths

Property of parallelogram: 2 parallel sides

Area = 1/2×4×(9+9)

        = 36 cm^2

3 0
2 years ago
Read 2 more answers
Translate the product of a number and 4 increased by 1 is -3 into a equation
laiz [17]
4x + 1 = -3

"The product" is the term used to describe the answer of a multiplication equation.
"increased by" means you're adding.
"a number" refers to a variable.
7 0
3 years ago
HELP ASAP!!!!!!!!
Hoochie [10]

Answer:

Step-by-step explanation:

The set {1,2,3,4,5,6} has a total of 6! permutations

a. Of those 6! permutations, 5!=120 begin with 1. So first 120 numbers would contain 1 as the unit digit.

b. The next 120, including the 124th, would begin with  '2'  

 c. Then of the 5! numbers beginning with 2, there are 4!=24 including the 124th number, which have the second digit =1

d. Of these 4! permutations beginning with 21, there are 3!=6 including the 124th permutation which have third digit 3

e. Among these 3! permutations beginning with 213, there are 2 numbers with the fourth digit =4 (121th & 122th), 2 with fourth digit 5 (numbers 123 & 124) and 2 with fourth digit 6 (numbers 125 and 126).

Lastly, of the 2! permutations beginning with 2135, there is one with 5th digit 4 (number 123) and one with 5 digit 6 (number 124).

∴   The 124th number is 213564

Similarly reversing the above procedure we can determine the position of 321546 to be 267th on the list.

8 0
2 years ago
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