MATHEMATICS HIGH SCHOOL

What is the highest common factor for 3876 and 114?

Answers

Answer 1

Answer:

The highest common factor would be 114

3876 = 2x3x3x17x19

114 = 2x3x19

Answer 2

Answer:

Final answer:

The highest common factor (HCF) of 3876 and 114 is 6.

Explanation:

To find the highest common factor (HCF) of two numbers, such as 3876 and 114, we need to determine the largest number that divides evenly into both given numbers.

We can start by finding the prime factors of each number. The prime factorization of 3876 is 2 × 3 × 1,29, and the prime factorization of 114 is 2 × 3 × 19.

The common factors we can identify are 2 and 3, so the HCF of 3876 and 114 is 2 × 3, which is 6.

Learn more about highest common factor here:

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What is the value of w in the equation 1/2w+7=2w-2

Answers

$(1)/(2)w+7=2w-2\ \ \ \ |subtract\ 7\ from\ both\ sides\n\n(1)/(2)w=2w-9\ \ \ \ |subtract\ 2w\ from\ both\ sides\n\n-1(1)/(2)w=-9\n\n-1.5w=-9\ \ \ \ |divide\ both\ sides\ by\ (-1.5)\n\n\boxed{w=6}$

Compare the decimals 18.10 and 18.1. Which comparison is correct? A. 18.10 = 18.1 B. 18.10 > 18.1 C. 18.10 < 18.1

Answers

18.10=18.1
The 0 in 18.10 does not affect the place value

Abbi invested $1,300 in a certificate of deposit with a simple interest rate of 3%. Find the interest earned in 6 years. Then find the total of principal plus interest.A.)$234.00; $1,534.00
B.)$23,400.00; $24,700.00
C.)$39.00; $1,339.00
D.)$252.27; $1,552.27

Answers

your answer is A..................................

These two functions represent the growth of two different bacterial cultures in terms of the number of bacteria after x days.f(x) = 2,000(2)^x (other one on the graph)

Answer the following:
A) Which function has the higher initial amount of bacteria? g(x) or f(x)
B) Which function has the greater amount of bacteria after two days? g(x) or f(x)

Answers

Answer:

A. The function f(x) has the higher initial amount of bacteria.

B. The function g(x) has the higher amount of bacteria after two days.

Step-by-step explanation:

The given function is

$f(x)=2000(2)^x$

The graph of g(x) passing through the points (0,1000) and (1,3000). So the initial value is 1000 and the growth factor is 3.

The function g(x) is

$g(x)=1000(3)^x$

Part A:

Substitute x=0, to find the initial blue of the functions.

$f(0)=2000(2)^0=2000$

$g(0)=1000(3)^0=1000$

Since 2000>1000, therefore the function f(x) has the higher initial amount of bacteria.

Part B:

Substitute x=2, to find the amount of bacteria after two days.

$f(2)=2000(2)^2=8000$

$g(2)=1000(3)^2=9000$

Since 8000<9000, therefore the function g(x) has the higher amount of bacteria after two days.

19x+(1/2)ax=x+9 solve for x

Answers

19x + ¹/₂ax = x + 9
- x - x
18x + ¹/₂ax = 9
x(18) + x(¹/₂a) = 9
x(18 + ¹/₂a) = 9
18 + ¹/₂a 18 + ¹/₂a
x = 9/(18 + ¹/₂a)

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

Answers

$\left( (27x^3)/(8y^9)\right)^ (5)/(3) \n\n\n =\left( ((3x)^3)/((2y^3)^3)\right)^ (5)/(3) \n\n\n = \frac{(3x)^{3 * (5)/(3) }}{(2y^3)^{3 * (5)/(3) }} \n\n\n =((3x)^5)/((2y^3)^(5 )) \n\n\n =(243x^5)/(32y^(15))$

Now, If the exponent was negative like you asked....

$\left( (27x^3)/(8y^9)\right)^ {-(5)/(3)} \n\n\n =\left( (8y^9)/(27x^3)\right)^ {(5)/(3)}\n\n\n =\left( ((2y^3)^3)/((3x)^3)\right)^ (5)/(3) \n\n\n = \frac{(2y^3)^{3 * (5)/(3) }}{(3x)^{3 * (5)/(3) }} \n\n\n =((2y^3)^(5 ))/((3x)^5) \n\n\n =(32y^(15))/(243x^5)$

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