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ololo11 [35]
3 years ago
5

Please Help! Solve for X

Mathematics
2 answers:
Juli2301 [7.4K]3 years ago
3 0

Answer:

x=2

I really don't know but I really tried

I really hope that this helps!

Vladimir79 [104]3 years ago
3 0
<h3>Given :</h3>

  • ax + by = lx - d

<h3>To Find :</h3>

  • Value of x = ?

<h3>Solution :</h3>

\tt \dashrightarrow ax + by = lx - d

\tt \dashrightarrow ax + by - lx + d= 0

\tt \dashrightarrow ax - lx + by + d= 0

\tt \dashrightarrow ax - lx = - by - d

By taking x common from (ax - lx), we get :

\tt \dashrightarrow x(a - l) = - by - d

\tt \dashrightarrow x = \dfrac{- by - d}{a-l}

Hence, values of \tt x = \dfrac{- by - d}{a-l}

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Given the formula an = 47 + (n - 1)(-3), find the 21st term in the sequence.
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7 0
3 years ago
Mrs. Gordon had $40 to spend on cookies and pies. She bought 4 cookies, each of which cost $3. She spent twice as much on pies a
DiKsa [7]
First, multiply $3 by 4 to get $12 spent on cookies.  Then, you multiply $12 by 2 to know that the pies cost $24.  You then divide the $24 by 3 to get $6 per pie.  Meaning, each pie had cost $6.  You add the $12 and $24 together to get $36.  Finally, you subtract $36 from $40 to know that she only has $4 left.
4 0
3 years ago
• Malik Montez's savings account has a principal of $1,640. It earns 6 percent interest a
jekas [21]

Answer:

The amount at the end of second quarter is $1842.704

Step-by-step explanation:

Given as :

The principal in saving account = $1640

The rate of interest = 6 % compounded quarterly

The time period = 6 months

Now,<u> from compounded method</u>

The amount at the end of first quarter = A_1 = principal + 6 % of principal

I.e  A_1 = $1640 + 6 % × $1640

Or, A_1 = $1640 + $ 98.4

∴ A_1 = $1738.4

Again ,

The amount at the end of second quarter = A_2 = A_ + 6 % of  A_1

I.e  A_2 = $1738.4 + 6 % × $1738.4

Or, A_2 = $1738.4 + $104.304

∴ A_2 = $1842.704

So, The amount at the end of second quarter =  A_2 = $1842.704

Hence , The amount at the end of second quarter is $1842.704 Answer

8 0
3 years ago
Determine the values of the constants B and C so that the function given below is differentiable.
laila [671]
For the function to be differentiable, its derivative has to exist everywhere, which means the derivative itself must be continuous. Differentiating gives

f'(x)=\begin{cases}24x^2&\text{for }x1\end{cases}

The question mark is a placeholder, and if the derivative is to be continuous, then the question mark will have the same value as the limit as x\to1 from either side.

\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}24x^2=24
\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}B=B

So the derivative will be continuous as long as B=24

For the function to be differentiable everywhere, we need to require that f(x) is itself continuous, which means the following limits should be the same:

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}8x^3=8
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}Bx+C=24+C

24+C=8\implies C=-16

So, the function should be

f(x)=\begin{cases}8x^3&\text{for }x\le1\\24x-16&\text{for }x>1\end{cases}

with derivative

f'(x)=\begin{cases}24x^2&\text{for }x
5 0
4 years ago
21.98 round to the nearest tenth
Molodets [167]
20.00

I hope this helps :)
5 0
3 years ago
Read 2 more answers
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