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3 years ago

Pls help will mark brainlist if correct !! :))

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

E. 31

Step-by-step explanation:

Add the values together.

4 (2/6) + 3 (3/6) + 2 (3/6) = 9 (8/6) or 10 (1/3)

multiply 10 by 3 to find how many thirds there are and add 1 to that number for the extra 1/3.

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The (greatest, least) of the common factors of 2 or more numbers is the (greatest, least) common factor of the numbers.

Liula [17]

Answer:

The greatest of the common factors of 2 or more numbers is the greatest common factor of the numbers.

Step-by-step explanation:

As we know that

The greatest common factor of two or more whole numbers is the largest whole number that divides evenly into each of the numbers.

Thus, the greatest of the common factors of 2 or more numbers is the greatest common factor of the numbers.

For example, lets find the greatest common factor of $36$ and $54$ .

The possible factors of $36$ are:

1, 2, 3, 4, 6, 9, 12, 18, and 36.

The possible factors of $54$ are:

1, 2, 3, 6, 9, 18, 27, and 54.

The common factors of 36 and 54 are:

$1, 2, 3, 6, 9, 18$

So, from the common factors, it is easy to figure out that 18 is the greatest common factor.

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3 years ago

Which expression is equivalent to the number 144?

motikmotik

Answer:

Step-by-step explanation:

4 0

2 years ago

A, B, and C are collinear B is between A and C.

mixas84 [53]

Answer:

x = 7

Step-by-step explanation:

A, B, and C are collinear B is between A and C.

AC = AB + BC

2x + 1 = x + 4 + 2x - 10

2x + 1 = 3x - 6

2x - 3x = - 6 - 1

- x = - 7

x = 7

5 0

3 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

<u>Taylor series</u> expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$