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natima [27]
3 years ago
13

Round 6.7086 to the nearest thousandth

Mathematics
1 answer:
Bingel [31]3 years ago
7 0

\text {Hello! Let's solve this Problem!}

\text {To round to the nearest thousandth you must underline the thousandth place}\text {(which is 8), point an arrow to 6 (in the ten-thousandth place)}

\text {If the number is 5 or more you raise the score. Meaning you add 1 to the}\text {thousandth place. If the number is 4 or lower you keep the number the same.}

\text {6 is greater than 5. So this means you add 1 to 8 and it becomes 9. Every}\text {number behind it becomes 0.}

\text {Your Answer Would Be:}

\huge\boxed {\text {6.7090 or 6.709}}

\text {Best of Luck!}

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What is the inverse of f(x)=3x+6 ?
scZoUnD [109]

Answer:

\boxed{\sf \ \ f^{-1}(x)=\dfrac{x-6}{3} \ \ }

Step-by-step explanation:

hello,

we can write

fof^{-1}(x)=x \ and \ fof^{-1}(x)=f(f^{-1}(x))=3f^{-1}(x)+6 \ so\\3f^{-1}(x)+6=x \ \ subtract \ 6\\3f^{-1}(x)=x-6 \ \ divide \ \ by \ \ 3\\ f^{-1}(x)=\dfrac{x-6}{3}

hope this helps

3 0
3 years ago
Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate
Harman [31]

Answer:

Yes

Step-by-step explanation:

Given that Machine M, working alone at its constant rate, produces x widgets every 4 minutes. Machine N, working alone at its constant rate, produces y widgets every 5 minutes.

When both machines work for 20 minutes

Machine A would produce = \frac{20}{4} =5 widget and

Machine B would produce = \frac{20}{5} =4 widgets

So we can say that Machine A would produce more widgets than Machine N at that time.

Answer is Yes

8 0
3 years ago
Please help don't understand how to do this, show steps
Anit [1.1K]
First get what equals 20
6 0
3 years ago
-7 2/3 + (-5 1/2) + 8 3/4=
myrzilka [38]
-4.41666666667



Here u go
3 0
2 years ago
Tom's stockbroker offers an investment that is compounded continuously at an annual interest rate of 3.7%. If Tom wants a return
Anettt [7]

Answer:

It'll take 38.3 years to obtain the desired return of $25,000.

Step-by-step explanation:

In order to solve a continuosly coumponded interest question we need to apply the correct formula that is given bellow:

M = C*e^(r*t)

Where M is the final value, C is the initial value, r is the interest rate and t is the time at which the money was applied. Since he wants an return of $25,000 his final value must be the sum of the initial value with the desired return. So we have:

(25000 + 8000) = 8000*e^(0.037*t)

33000 = 8000*e^(0.037*t)

e^(0.037*t) = 33000/8000

e^(0.037*t) = 4.125

ln[e^(0.037*t)] = ln(4.125)

t = ln(4.125)/(0.037)

t = 1.4171/0.037 = 38.2991

t = 38.3 years

7 0
3 years ago
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