Solve the linear equation 5(8x+2) -64=2(8x9) Find X

PLEASE HELP, MATH, PLEASE<br>Solve for n: n+5/16 = -1

Anit [1.1K]

$\tt Step-by-step~explanation:$

To solve for n, we have to isolate n. To do so, we move all the terms that are not n to one side of the equation, and leave n on the other side.

$\tt Steps:$

Equation: n + 5/16 = -1

Subtract 5/16 on both sides to bring it to the right side of the equation.

$\tt n+5/16-5/16=-1-5/16\\\\n=-\frac{21}{16}~or~-1\frac{5}{16}$

$\Large\boxed{\tt Our~final~answer:~n=-\frac{21}{16}~or~-1\frac{5}{16} }$

3 0

2 years ago

On his first day of school, Kareem found the high temperature in degrees Fahrenheit to be 76.1°. He plans to use the function to

wolverine [178]

The function for conversions from Fahrenheit to Celsius is:

Celsius = (Fahrenheit - 32) x 5/9

C(76.1) = (76.1 - 32) x 5/9
= (44.1) x 5/9
= 24.5

Therefore, 76.1 degrees Fahrenheit = 24.5 degrees Celsius.

5 0

3 years ago

Read 2 more answers

Sandra is opening a savings account which compounds interest quarterly. Her banker gave her the following function to find the a

anygoal [31]

A ..... Because .....

4 0

3 years ago

Math question down below

julsineya [31]

This data set has no outliers, because all of the numbers are close together.

6 0

2 years ago

Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5

alex41 [277]

The smallest such number is 1055.

We want to find $x$ such that

$\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}$

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

$x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}$

$x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}$

Taking everything together, we end up with the system

$\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}$

Now the moduli are coprime and we can apply the CRT.

We start with

$x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5$

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2$

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

$x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3$

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

$x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5$

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

$x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7$

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since $2\cdot4\equiv8\equiv1\pmod7$ , the inverse of 2 is 4.

So, we have to adjust $x$ to

$x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845$

and from the CRT we find

$x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}$

so that the general solution $x=210n+5$ for all integers $n$ .

We want a 4 digit solution, so we want

$210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5$

which gives $x=210\cdot5+5=1055$ .

5 0

2 years ago