In the diagram, JG = 5 cm and GE = 10 cm. Based on this information, can G be a centroid of triangle HJK? Point G cannot be a centroid because JG does not equal GE. Point G cannot be a centroid because JG is shorter than GE. Point G can be a centroid because GE and JG are in the ratio 2:1. Point G can be a centroid because JG + GE = JE.

Answer:

Another word for mean is the "average", your goal is to find the average of the following data set, it's pretty easy, you just have to add up all the numbers in the data set and divide it with the total numbers in the data set.

For example:

Find the mean of the following data set:

4, 1, 8, 5, 7, 3

Mean: 4 + 1 + 8 + 5 + 7 + 3 = 28

then we divide 28 with 6 since there are 6 numbers in the following data set.

Mean: 4 + 1 + 8 + 5 + 7 3 = 28/6 = 4.6

We get the final answer of 4.6!

Answer:

250

Step-by-step explanation:

Answer: Second option is correct.

Step-by-step explanation:

Since we have given that

Amount for policy = $50,000

If he opt for life insurance @$5.84 per $1000,

then, his amount of premium will be

[tex]50000\times \frac{5.84}{1000}=\$292[/tex]

If he opt for straight policy @ $15.66 per $1000,

then, his amount of premium will be

[tex]50000\times \frac{15.66}{1000}=\$783[/tex]

Difference in both the cases will be

[tex]783-292=\$491[/tex]

Hence, Second option is correct.

I KNOW ITS NOT A. But i think it is B. HOPE THIS HELPS!

Answer:

[tex]10\ stamp\ packets[/tex]

Step-by-step explanation:

Let

x-----> the number of packets that mark can buy

we know that

[tex]1.78x\leq 19[/tex] -----> inequality that represent the situation

Solve for x

Divide by [tex]1.78[/tex] both sides

[tex]x\leq 19/1.78[/tex]

[tex]x\leq 10.67\ stamp\ packets[/tex]

The domain of the composite function (cd)(x), where c(x) = 5/(x-2) and d(x) = x + 3, includes all real values except x = 2 due to a division by zero condition in the function c(x).

In this problem, we need to figure out the domain of the combined function (cd)(x). The domain constitutes the set of all real values of x that make the function defined. The first function, c(x) =  rac{5}{x-2}, isn't defined for x = 2 because it would lead to a division by zero. The second function, d(x) = x + 3, is defined for all real values since it's a linear function.

When these functions are multiplied to form (cd)(x), the domain will be the intersection of the individual function domains. This means, with c(x) having a restriction with x != 2, the composite function (cd)(x) will maintain this restriction despite d(x) being defined for all real values. So the domain for (cd)(x) is all real values except for x = 2.

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Answer:

43.755 what the other person said but I know that is the answer

Step-by-step explanation:

Answer:

false 9.13 is greater

Step-by-step explanation:

103 thousanths are smaller then 13 hundreths

with decimals the farther you go the lower you get bc its half a number

1/2 to 1/4 to  1/8 it gets smaller

Answer: 30

Explanation: To find the least common multiple of lcm of 5 and 6, we begin by listing the first few multiples of each number.

Multiples of 5

1 x 5 = 5

2 x 5 = 10

3 x 5 = 15

4 x 5 = 20

5 x 5 = 25

6 x 5 = 30

Notice that we skipped 0 x 6 in our list of multiples. That's because 0 x 6 is 0 and our least common multiple can't be 0.

Next we list the multiples of 6. When we list the multiples of 6, it's a good idea to keep an eye on the list of multiples for 6 so that we know when we have found a least common multiple.

Multiples of 6

1 x 6 = 6

2 x 6 = 12

3 x 6 = 18

4 x 6 = 24

5 x 6 = 30

Notice that 30 appears in both lists so we can stop here.

This means that the least common multiple of 5 and 6 is 30.

Answer:

30

Step-by-step explanation:

both 6 and 5 go into 30 :)

Answer:

42 125 becuase i got um right bozzzzzzzz

Step-by-step explanation:

Answer:

Option A, C, D, are the correct options.

Step-by-step explanation:

To check the exponential functions which have been simplified correctly we will take option one by one.

A). [tex]f(x)=5(\sqrt[3]{16})^{x}=5(\sqrt[3]{2^{4}})^{x}=5(2\sqrt[3]{2})^{x}[/tex]

B). [tex]f(x)=2.3(8^{\frac{1}{2}})^{x}=2.3(2.\sqrt{2})^{x}\neq 2.3(4^{x})[/tex]

C). [tex]f(x)=81^{\frac{x}{4}}=(3^{4})^{\frac{x}{4}}=(3^{\frac{4}{4}})^{x}=3^{x}[/tex]

D). [tex]f(x)=\frac{3}{4}(\sqrt{27})^{x}=\frac{3}{4}(\sqrt{3^{3}})^{x}=\frac{3}{4}.(3\sqrt{3})^{x}[/tex]

[tex]f(x)=(24)^{\frac{1}{3}x}=(2^{3}.3)^{\frac{1}{3}x}=(2.3^{\frac{1}{3}})^{x}\neq 2(\sqrt[3]{3})^{x}[/tex]

Exponential functions and their inverses like the natural logarithm are inverse functions, meaning they can undo each other. Simplifications like ln(e^x) result in x, and e^(ln x) simplifies to x. This property is integral to correctly simplifying exponential functions.

The exponential functions and their inverses, like the natural logarithm (ln), have specific properties where they can "undo" each other. This relationship allows us to simplify expressions involving exponentials and logarithms. For example, when you have ln(ex), this simplifies to x because the natural logarithm and the exponential function are inverse functions.

Similarly, when applying the exponential function to a natural logarithm, like eln(x), the result is just x. Furthermore, when taking square roots of exponentials, you can adjust the exponent to be evenly divisible by 2 and then take the square root of the base number, followed by dividing the exponent by 2. Also, while dividing exponentials, you subtract the exponents when the bases are the same.

Here is an example of this simplification: to compute the square root of e4, we can write it as e4/2, which simplifies to e2.