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its hammer time

Log ₁₀ x  = 7

Exponentially, this says that

10⁷  =  x  =  10,000,000

No more May the 4th be with you.

Chad built a scale model of a statue.

He built the model 7 inches tall to represent the actual height of 15 feet.

Which equation below represents the relationship between the actual height (a) ,in feet and the height of the model (m) , in inches?

$a\;\; \alpha\;\; m\\ a\;\; =\;\;k * m\\ \text{When actual is 15 model is 7}\\ 15\;\; =\;\;k * 7\\ k=\frac{15}{7}\\ a\;\; =\;\;\frac{15}{7} * m\\ a\;\; =\;\;\frac{15m}{7} \\ $

X=.7

If you multiply that by 10, it equals 7.

May the 4th  be with you!!

im good thanks

How I should resolve (1/(2))^x=16 without the calculator?

$\begin{array}{|rcll|} \hline \left(\frac12 \right)^x &=& 16 \\ \left(\frac12 \right)^x &=& 2^4 \\ \left(\frac12 \right)^x &=& \frac{1}{2^{-4}} \\ \left(\frac12 \right)^x &=& \left(\frac12 \right)^{-4} \\ x &=& -4 \\ \hline \end{array}$

you got it right

go girl go

In the case of sums, that could be a combonation of two numbers from 0 to 122.

0+122

1+121

2+120

3+119

and on.

What time?

Time taken for heating it to 100 degrees celcius? Or time taken for pouring it into a pool?

Can you explain what is the term "time" for?

You did the similar things as I did :P

There is insufficient information in the question,

the quadratic  $\displaystyle f(x) = 24x^{2}-192x +396$

fits the sequence just as well.

You need to be told what sort of function that you're looking for.

(It's also possible to find an infinite number of cubics, quartics ... etc. as well.)

I'm guessing a bit here, but how about the following:

Suppose your numbers are x₁, x₂ and x₃

Calculate the mean:  $x_{av} = (x_1+x_2+x_3)/3$

Calculate the standard deviation:   $s = \sum_{k=1}^3(x_k-x_{av})^2/2$

Calculate $score_k = 2\frac{x_k-x_{av}}{s}+5$ for k = 1, 2 and 3

Then score will have a mean of 5 and a standard deviation of 2.  

Round the resulting values to the nearest integer to get the stanine values.

As I said above, I'm guessing here!

If you want to know what sequence the numbers form, then it is as follows:

4, 12, 5, 36, 6, 108, 7, 324, 8, 972.........etc.

First term goes up by 1, such as 4+1, 5+1, 6+1......etc.

Second term is multiplied by 3, such as 12 x 3, 36 x 3, 108 x 3 .......and so on.

"how do you find the apothem of a square"

It's half the length of a side.

.

what question?

Need help with this continuous fraction integral.

integral \limits_{x=0}^1  \frac{x\pm\sqrt{x^2+4x} }{2}  \ dx

Continuous Fraction.

$\begin{array}{|rcll|} \hline \begin{equation*} sum=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{x+\cdots}}} \end{equation*}\\\\ sum &=& x + \frac{x}{sum} \\ sum - \frac{x}{sum} &=& x \\ \frac{sum^2-x}{sum} &=& x \\ sum^2-x &=& x\cdot sum \\ sum^2-x\cdot sum -x &=& 0 \\ sum &=& \frac{x\pm\sqrt{x^2-4\cdot(-x)} }{2} \\ \mathbf{sum} & \mathbf{=} & \mathbf{ \frac{x\pm\sqrt{x^2+4x} }{2} } \\ \hline \end{array}$

$\small{ \begin{array}{llcl} \int \limits_{x=0}^{1} { x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{x+\cdots}}} \ dx} \\\\ = \int \limits_{x=0}^{1} { sum \ dx} \\ = \int \limits_{x=0}^{1} { \mathbf{ \frac{x\pm\sqrt{x^2+4x} }{2} } \ dx} \\ = \frac12 \int \limits_{x=0}^{1} {x \ dx } \pm \frac12 \int \limits_{x=0}^{1} { \sqrt{x^2+4x} \ dx } \\ = \frac12 \left[\frac{x^2}{2}\right]_{x=0}^{1} \pm \frac12 \int \limits_{x=0}^{1}{ \sqrt{(x+2)^2-4} \ dx } \\ = \frac14 \pm \frac12 \int \limits_{x=0}^{1}{ \sqrt{(x+2)^2-4} \ dx } \\ & \text{substitute:}\\ & \boxed{~ u=x+2\\ du = dx ~}\\ & \text{new limits:}\\ & \boxed{~ x=0: \qquad u=0+2 \Rightarrow u = 2 \\ x=1: \qquad u=1+2 \Rightarrow u = 3 ~}\\ = \frac14 \pm \frac12 \int \limits_{u=2}^{3} { \sqrt{u^2-4} \ du } \\ & \text{substitute:}\\ & \boxed{~ u=2 \cosh(z) \qquad z = \text{arcosh} \left(\frac{u}{2}\right)\\ du = 2 \sinh(z)\ dz ~}\\ & \text{new limits:}\\ & \boxed{~ u=2: \qquad z=\text{arcosh} \left(\frac{2}{2}\right) \\ \Rightarrow z = \text{arcosh}(1) = 0 \\ u=3: \qquad z=\text{arcosh} \left(\frac{3}{2}\right) \\ \Rightarrow z = \text{arcosh}(1.5) ~}\\ = \frac14 \pm \frac12 \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\cosh(z)^2-4} \cdot 2 \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\cosh(z)^2-4} \cdot \sinh(z) \ dz } \\ & \boxed{~ \cosh^2(z) - \sinh^2(z) = 1 \\ \cosh^2(z) - 1 = \sinh^2(z) \quad | \quad \cdot 4 \\ 4\cosh^2(z) - 4 = 4\sinh^2(z) \\ ~}\\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \sqrt{4\sinh^2(z)} \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { 2\sinh(z) \cdot \sinh(z) \ dz } \\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { 2\sinh^2(z) \ dz } \\ \end{array} }$

$\small{ \begin{array}{llcl} & \boxed{~ \cosh(2z) = \cosh^2(z) + \sinh^2(z) \\ \cosh(2z) = 1+\sinh^2(z) + \sinh^2(z) \\ \cosh(2z) = 1+2\sinh^2(z) \\ 2\sinh^2(z) = \cosh(2z)-1 ~}\\ = \frac14 \pm \int \limits_{z=0}^{\text{arcosh}(1.5)} { \Big( \cosh(2z)-1 \Big) \ dz } \\ = \frac14 \pm \Big( \int \limits_{z=0}^{\text{arcosh}(1.5)} {\cosh(2z) \ dz } -\int \limits_{z=0}^{\text{arcosh}(1.5)} {\ dz } \Big) \\ = \frac14 \pm \Big( \left[\frac12\cdot \sinh(2z) \right]_{z=0}^{\text{arcosh}(1.5)} -\left[ z \right]_{z=0}^{\text{arcosh}(1.5)} \Big) \\ & \boxed{~ \sinh(2z) = 2\sinh(z)\cosh(z) ~}\\ = \frac14 \pm \Big( \left[\frac12\cdot 2\sinh(z)\cosh(z) \right]_{z=0}^{\text{arcosh}(1.5)} - \text{arcosh}(1.5) \Big) \\ = \frac14 \pm \Big( \left[ \sinh(z)\cosh(z) \right]_{z=0}^{\text{arcosh}(1.5)} - \text{arcosh}(1.5) \Big) \\ = \frac14 \pm \Big[ \sinh\Big(\text{arcosh}(1.5)\Big)\cdot 1.5 - \text{arcosh}(1.5) \Big] \\ & \boxed{~ \cosh^2(x) - \sinh^2(x) = 1 \\ \sinh^2(x) = \cosh^2(x) - 1 \\ \sinh(x) = \sqrt{\cosh^2(x) - 1} \\ ~}\\ \boxed{~ \sinh\Big(\text{arcosh}(x)\Big)= \sqrt{\cosh^2(\text{arcosh}(x)) - 1} = \sqrt{x^2 - 1} \\ \sinh\Big(\text{arcosh}(1.5)\Big) = \sqrt{1.5^2 - 1} = \sqrt{1.25} = \frac{ \sqrt{5} }{2} ~}\\ = \frac14 \pm \Big[ \frac{\sqrt{5}}{2} \cdot \frac{3}{2} - \text{arcosh}(1.5) \Big] \\ = \frac14 \pm \Big[ \frac{3\sqrt{5}}{4} - \text{arcosh}(1.5) \Big] \\ = \frac14 \pm ( 1.6770509831248424 - 0.9624236501192069 ) \\ = \frac14 \pm 0.7146273330056354 \\ \end{array} }$

$\begin{array}{|rcll|} \hline = \frac14 + 0.7146273330056354 \qquad &\text{or}&\qquad = \frac14 - 0.7146273330056354 \\ = 0.9646273330056354 \qquad &\text{or}& \qquad =-0.4646273330056354 \\ \hline \end{array}$

If you just want the answer for    $\tan(\frac12x-20º)$  in the first quadrant...

$\tan(\frac12x-20º)=1\\~\\ \frac12x-20º=\arctan(1) \\~\\ \frac12x-20º=45º \\~\\ \frac12x=45º+20º \\~\\ \frac12x=65º \\~\\ x=65º*2\\~\\x=130º$

If you want the general solution please say so 

I think you want the equation of a line that passes through the point (2, -2) and has a slope of 2/7 .

y - y₁ = m(x - x₁)

y₁ = -2     and     x₁ = 2

Plug in.

y - (-2) = (2/7)(x - 2)

Solve for y and simplify.

y = (2/7)(x - 2) - 2

y = (2/7)x - (2/7)(2) - 2

y = (2/7)x - 4/7 - 14/7

y = (2/7)x -18/7