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3 years ago

13

Rayhana has a coupon for 15% off a single item at a store. If the item has a regular price of p dollars, then the sale price can

be represented by p – 0.15p. Which expression is equivalent?

2 answers:

IgorC [24]3 years ago

5 0

0.85pis the equivalent expression

Sergeu [11.5K]3 years ago

4 0

Answer:

$0.85p$

Step-by-step explanation:

We have been given an expression $p-0.15p$ , which represents the price of an item p after a 15% off. We are asked to find equivalent expression.

We can find equivalent expression in two ways. Either by finding 85% of p (100%-15%) or factoring our p from our given expression as:

$(1-0.15)p$

$0.85p$

85% of p: $\frac{85}{100}p=0.85p$

Therefore, the expression $0.85p$ is equivalent expression to our given expression.

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Using a calcuator, $7.6^{2} =57.76$

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" Pentagon ABCDE and pentagon A’B’C’D’E’ are shown on the coordinate plane below.

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3 years ago

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An automotive repair center charges $45 for any part of the first hour of labor, and $25 for any part of each additional hour. W

IceJOKER [234]

The total cost is given by the equation:
C(t) = 45 + 25(h-1) where h is the number of hours worked.

We can check for each option in turn:

Option A:

Inequality 5 < x ≤ 6 means the hour is between 5 hours (not inclusive) to 6 hours (inclusive)
Let's take the number of hours = 5
C(5) = 45 + (5-1)×25 = 145
Let's take the number of hours = 6
Then substitute into C(6) = 45 + (6-1)×25 = 170
We can't take 145 because the value '5' was not inclusive.

Option B:
The inequality is 6 < x ≤ 7
We take number of hours = 6
C(6) = 25(6-1) + 45 = 170
We take number of hours = 7
Then C(7) = 25(7-1) + 45 = 195

Option C:
The inequality is 5 < x ≤ 6
Take the number of hours = 5
C(5) = 25(5-1) + 45 = 145
Take the number of hours = 6
C(6) = 25(6-1) + 45 = 170
We can't take the value 145 as '5' was not inclusive in the range, but we can take 170

Option D:
6 < x ≤ 7
25(6-1) + 45 < C(t) ≤ 25(7-1) + 45
170 < C(t) ≤ 195

Correct answer: C

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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

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

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3 years ago

el vector desplazamiento de un objeto en movimiento inicialmente en el origen tiene una magnitud de 12.5 cm y está en un ángulo

vladimir1956 [14]

Answer:

wait what? sorry I don't understand .

7 0

3 years ago

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