All Answers 357329 Answers

(2,106.67/223,660,000,000)*100?

$(\frac{2106.67}{223660000000})*100$

Using the rule of PEMDAS, we have to simplify inside the parenthesis first.

$(0.0000000094190736)*100$

Now multiply.

$0.0000000094190736*100 = 0.00000094190736​$

Logorithm

It is it's own formula, if you were to use the "LOG" button, the calculator would input the standardized formula and give you an output.

Wow, we got some smart people on here! This math problem was not mine but instead from a math program called ACT. 

Great job hectictar! You got it right!

Here's the graph : https://www.desmos.com/calculator/tew5gj5h5z

Note that -3/2  =  -1.5

To graph this point, start at the origin (0,0)

- 1.5 means that we then  move to the left 1.5  units.......then....go up 3 units......this will be your point, (-3/2 , 3)

We have nothing to solve.....there needs to be an equal sign  somewhere for that

Domain  = number of possible cases he could buy  =  0  to 8

Range. =   nubmer of possible bottles of water he could have  =

7   to   [ 24(8)  + 7 ]  =     7 to  199

Domain  = number of possible packages he could buy =  0 to 9

Range  =  number of possible rolls  =

5  to  [ 12(9) + 5 ]   =     5 to  113

f (3x - 1)  =  f (5)     which implies that

3x - 1  =  5      add 1 to both sides

3x = 6     divide both sides by 3

x = 2

So.....subbing this value into the function, we have

(2)^2  + (2)  + 1     =      4  + 2  +  1    =     7 

 49 = 4a+9    subtract  9 from both sides

40  =  4a   divide both sides by 4

10  =  a

Q (x)  = ( 4x^3 - 2x^2 + 7x - 1)^2

Every term in Q(x)  will have an x associated with it except the last one which is "1"

And the y intercept of the function will have an associated x value of 0

So......evey term will evaluate to 0  except the last one

So.....the y intercept of Q(x)  =  1

If your numbers are all integers then the answer is no, because sqrt(3) is irrational.  If they can be reals, then yes:  (sqrt(3)*a)² = b², for which there are an infinite number of solutions.

Thank you so much! This is very helpful information!

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First expand the bracketed terms to get:

(m-5)(m+4) + 8 → m² - m - 20 + 8 → m² - m - 12

Now notice that 12 is the product of 4 and 3, and 1 is the difference between 4 and 3.

Taking the signs into account this gives us: m² - m - 12 = (m - 4)(m + 3)

I don't think so, sorry!

exponential problem

4^(x) -2^(x+2) +4=0

$\begin{array}{|rcll|} \hline 4^{x} -2^{x+2} +4 &=& 0 \\ (2\cdot2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \\ (2^2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \quad & | \quad (2^2)^{x} = 2^{2x} = 2^{x\cdot 2}= (2^x)^2 \\ (2^x)^2 -2^x\cdot2\cdot 2 +2^2 &=& 0 \quad & | \quad \text{binom } \\ (2^x-2)^2 &=& 0 \quad & | \quad \text{square root both sides}\\ 2^x-2 &=& 0 \quad & | \quad +2 \\ 2^x &=& 2 \quad & | \quad 2=2^1 \\ 2^x &=& 2^1 \\ \mathbf{x} &\mathbf{=}& \mathbf{ 1} \\ \hline \end{array}$

Find the value(s) of m in the quadratic x²+x+m+1=0
such that there is only one solution.

$\begin{array}{|lrcll|} \hline & x^2+1\cdot x+ (m+1) &=& 0 \\ & \text{ or } (x-x_1) (x-x_2) &=& 0 \quad & | \quad \text{one solution so } x_1 = x_2 \\ & (x-x_1) (x-x_1) &=& 0 \\ & (x-x_1)^2 &=& 0 \\ & x^2 \underbrace{-2x_1}_{=1}x +\underbrace{ x_1^2}_{=m+1} &=& 0 \quad & | \quad \text{compare the coefficients} \\\\ (1) & -2x_1 &=& 1 \\ & x_1 &=& -\frac{1}{2} \\\\ (2) & x_1^2 &=& m+1 \\ & \left(-\frac{1}{2}\right)^2 &=& m+1 \\ & \frac{1}{4} &=& m+1 \\ & \frac{1}{4}-1 &=& m \\ & \frac{1}{4}-\frac{4}{4} &=& m \\\\ &\mathbf{ -\frac{3}{4}} &\mathbf{=}& \mathbf{m} \\ \hline \end{array}$

Courtesy of Heureka :)

$\begin{array}{|rcll|} \hline && \mathbf{(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8 } \\ &=& (t-1)^3 + 3(t-1)^2+3(t-1) +1 \\ && \quad + 3(t-1)^2+9(t-1)+7 \\ &=& t^3 + 3(t-1)^2+9(t-1)+7 \\ &=& t^3 + 3(t^2-2t+1)+9t-9+7 \\ &=& t^3 + 3t^2-6t+3+9t-9+7 \\ &=& t^3 + 3t^2+3t+1 \quad & | \quad (1+t)^3 = t^3+3t^2+3t+1 \\ &\mathbf{=}& \mathbf{(1+t)^3} \\ \hline \end{array}$

\begin{array}{|rcll|} \hline && \mathbf{(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8 } \\

&=& (t-1)^3 + 3(t-1)^2+3(t-1) +1\\

 && \quad + 3(t-1)^2+9(t-1)+7 \\

&=& t^3 + 3(t-1)^2+9(t-1)+7 \\

&=& t^3 + 3(t^2-2t+1)+9t-9+7 \\

&=& t^3 + 3t^2-6t+3+9t-9+7 \\

&=& t^3 + 3t^2+3t+1 \quad & | \quad (1+t)^3 = t^3+3t^2+3t+1\\

&\mathbf{=}& \mathbf{(1+t)^3} \\

\hline \end{array}

I think by looking at the graph, (and using Melody's hint) the answer might be E.

Because the x-values are between 1 and 2 when y=(x-1)^4 is beneath y=x-1.

Find the value(s) of m in the quadratic x2+x+m+1=0 such that there is only one solution.

This appears to be correct, so Mr.Owl has met his/her match.

Use exactly the same formula and procedure as in your first question. You must change the numbers !!!!

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