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2 years ago

PLS HELP ASAP PLSSS IT DETECTS IF ITS RIGHT OR WRONG HELPPP

Mathematics

1 answer:

timurjin [86]2 years ago

3 0

Answer:

y= 250 - 2x

Step-by-step explanation:

She spends $2 per bottle, which is my it's 2x.

She has a total of 250, and she spends $2 per bottle, so you subtract 2x from 250, which leaves the equation at y = 250 - 2x

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Lydia graphed ΔXYZ at the coordinates X (0, −4), Y (2, −3), and Z (2, −6). She thinks ΔXYZ is a right triangle. Is Lydia's asser

katrin [286]

Answer:

Option (4)

Step-by-step explanation:

Vertices of the triangle are X(0, -4), Y(2, -3) and Z(2, -6).

Slope of XY ( $m_{1}$ ) = $\frac{y-y'}{x-x'}$

= $\frac{-3+4}{2-0}$

( $m_{1}$ ) = $\frac{1}{2}$

Similarly, slope of XZ ( $m_{2}$ ) = $\frac{-6+4}{2-0}$

( $m_{2}$ ) = (-1)

$m_{1}\times m_{2}=-\frac{1}{2}$

Which should be (-1) if XY and XZ are perpendicular to each other.

Now we can say that XY and XZ are not perpendicular.

Therefore, Lydia's assertion that ΔXYZ is a right triangle is not correct.

Option (4) will be the answer.

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2 years ago

Two friends, Devon and Bryce, are working together at the Oakland Cafe today. Devon works every 2 days, and Bryce works every 9

ludmilkaskok [199]

Answer:

11 days before they worked together

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

Find the order pairs for the x- and y- intercepts of the equation 2x-6y=24

dolphi86 [110]

The answer is going to be
x= 2,0
y= 0, -6

5 0

3 years ago

The table shows how an elevator 500 feet above the ground is descending at a steady rate.

lapo4ka [179]

Answer:

h(t) = -5t + 500

Step-by-step explanation:

At time zero, the elevator is at 500 ft.

At time 5 seconds, the elevator is at 475 ft.

The difference in height is -25 ft.

The elevator traveled -25 ft (25 ft down) in 5 seconds, so it travels at a speed of 5 ft per second. Since it is traveling down, its speed is -5 ft/sec.

For each second of travel, it loses 5 ft of height.

h(t) = -5t + 500

7 0

2 years ago