Why is using coins as money easier than using gold bars?

Mathematics

1 answer:

Solnce55 [7]3 years ago

5 0

Answer:

because coins are more portable

Step-by-step explanation:

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Please help how to solve

tatiyna

Answer:

$= > in \: the \: bigger \: triangle \\ height = 14 \: ft \\ base = 22.5 \: ft \\ = > in \: smaller \: triangle \\ height = x \: ft \\ base = 9 \: ft \\ using \: ratios \\ \frac{bigger \: height}{smaller \: height} = \frac{bigger \: base}{smaller \: base} \\ \frac{14}{x} = \frac{22.5}{9} \\ \frac{14}{x} = 2.5 \\ x = \frac{14}{2.5} \\ x = 5.6 \: ft$

7 0

3 years ago

After being rearranged and simplified, which of the following equations could be solved using the quadratic formula?

dmitriy555 [2]

A. 2x^2 - 3x + 10 = 2x + 21
2x^2 - 3x - 2x + 10 - 21 = 0
2x^2 - 5x - 11 = 0 (quadratic equation)

2x^2 - 6x - 7 = 2x^2
2x^2 - 2x^2 - 6x - 7 = 0
-6x - 7 = 0 (not a quadratic equation)

5x^2 + 2x - 4 = 2x^2
5x^2 - 2x^2 + 2x - 4 = 0
3x^2 + 2x - 4 (quadratic equation)

5x^3 - 3x + 10 = 2x^2
5x^3 - 2x^2 - 3x + 10 = 0 (not a quadratic equation)

Therefore, options a and c can be solved using the quadratic formula.

3 0

3 years ago

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$