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3 years ago

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Give one example of an equation with variables on both sides that has all real numbers as the solution. Also give an example of

an equation with variables on both sides that has no real solutions.

1 answer:

Ghella [55]3 years ago

6 0

1) 2x + 3 = 2x + 3
This works because if you solve it, you will get either x = x or 3 = 3 which are always true, so x, can have an infinite number of solutions
2) 3(x + 4) = 3x + 11
This has no solution because if you solve it, you're gonna get 12 = 11 and that is NEVER true. Whatever x is, 11 cannot equal 12!

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(4t+8)+t=414
Blah blah, I don't feel like doing work so I'll just use process of deduction and put the right answer here
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3 years ago

The amplitude of y = -2 sin 3x is

Rzqust [24]

Answer:2

Step-by-step explanation:

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3 years ago

The lines on a 2 cup liquid measuring cup divide each cup into eighths. If you measure 1 cups of water, between

Lemur [1.5K]

Answer:

The 4rth line and 5th line

Step-by-step explanation:

This 2 cup measuring cup with 8 division lines

8th line ------ 2 cups

7th line ------

6th line ------

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6 0

3 years ago

Read 2 more answers

What is the probability that a point chosen at random in the given figure will be inside the larger square and outside the small

liraira [26]

Answer:

51/100

Step-by-step explanation:

<em>The probability that a point chosen at random in the given figure will be inside the larger square and outside the smaller square</em> is equal to the ratio of the area of interest to the total area:

<em>P(inside larger square and outside smaller square)</em> = area of interest / total area

<em>P(inside larger square and outside smaller square)</em> = area inside the larger square and outside the smaller square / area of the larger square

<u>Calculations:</u>

<u />

1. <u>Area inside the larger square</u>: side² = (10 cm)² = 100 cm²

2. <u>Area inside the smaller square </u>= side² = (7cm)² = 49 cm²

3. <u>Area inside the larger square and outside the smaller square</u>

100 cm² - 49 cm² = 51 cm²

4.<u> P (inside larger square and outside smaller squere)</u>

51 cm² / 100 cm² = 51/100

5 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

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

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

$a_{\boldsymbol{N}}=1 (2)^2=4$

3 0

3 years ago