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3 years ago

6.6666 to the thousands

1 answer:

8 0

6.6666 rounded to the nearest thousandth is 6.667.
The number in the ten thousandths place is greater than five, so the 6 in the thousandths place rounds up to a seven

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The cost of 4 pounds of bananas is $3.52. What is the constant of proportionality that relates the cost of dollars, y, to the nu

Sladkaya [172]

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3.52 divided by 4

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3 years ago

What is 347,168 round to then thoudands

Finger [1]

Answer:

347,168 rounded to the nearest ten thousands would be 350,000.

Step-by-step explanation:

When rounding to the nearest ten thousands, you rely on the thousands place value to indicate whether to round up or down. 5 or higher give it a shove, 4 or less let it rest. 7 is in the thousands place meaning we round up. 4 is in the ten thousands place meaning it becomes 5. 347,168 rounded to the nearest ten thousands would be 350,000.

8 0

3 years ago

Read 2 more answers

Find the solution(s) of 6x – 9 = 33x

Nookie1986 [14]

Answer:

C. x = 3/13

Step-by-step explanation:

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20%2B%201%20%7D%7B2%7D%20%20%3D%20%20%5Cfrac%7B7x%20-%203%7D%7B2%7D%20" id="Tex

MAVERICK [17]

Answer:

$\sf x =\dfrac{2}{3}$

Explanation:

$\sf \rightarrow \dfrac{x+1}{2} = \dfrac{7x-3}{2}$

cross multiply

$\sf \rightarrow 2(x+1) = 2(7x-3)$

divide both sides by 2

$\sf \rightarrow x + 1 = 7x-3$

isolate variables

$\sf \rightarrow x -7x = -3-1$

simplify

$\sf \rightarrow -6x = -4$

divide both sides by -6

$\sf \rightarrow \dfrac{-6x}{-6} = \dfrac{-4}{-6}$

simplify

$\sf \rightarrow x =\dfrac{2}{3}$

4 0

1 year ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago