Ask question

Ask question

All categories

4 years ago

6

adriana plans to use $260 less than three-fourths of her savings to buy a car. If the purchase price of the car is $9340, how mu

ch does she have in savings?

2 answers:

Alisiya [41]4 years ago

4 0

Let x = Adriana's savings
(3/4)x - 260 = 9340
(3/4)x = 9600
x = $12800

frozen [14]4 years ago

3 0

This one is easy, but tricky!

Let's say her total savings = x (4/4)

So, 9340 = (3/4x) - 260
9340 + 260 = 3/4x
9600 = 3/4x
(9600x4)/3 = x
12800 = x

You might be interested in

At cheng’s bike rental cost $40 to rent a bike for seven hours how many dollars does it cost per hour of bike use

Alchen [17]

40$. 7hrs
X$. 1hr
X=40•1/7
X~5.71 $ per hour

3 0

3 years ago

PLEASE HELP ME WITH THIS QUESTION ASAP. please and thank you

kotegsom [21]

To solve the given equation, you would need to multiply t by both terms inside the parenthesis.

The equation would be D. (t*14) - (t*5)

3 0

4 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

If the flare is launched with an initial velocity of 144 feet per second, find the height after 2 seconds. The answer is not 288

Rashid [163]

H= ut -16 t^2

h = 144(2) - 16 (4)

= 288 - 64 = 224

4 0

3 years ago

. A special safe lets you choose from 9 symbols for a 3-symbol-long pass code. You may enter a symbol any number of times. How m

Lady_Fox [76]

The question is an illustration of combination and there are 729 potential pass codes available

<h3>How to determine the number of potential pass codes?</h3>

The given parameters are

Symbols available = 9

Length of pass code = 3

From the question, we understand that a symbol may be entered any number of times.

This means that each of the 9 available symbols can be used three times

So, the number of potential pass codes is

Passcodes = 9 * 9 * 9

Evaluate the product

Passcodes = 729

Hence, there are 729 potential pass codes available

Read more about combination at:

brainly.com/question/11732255

#SPJ1

4 0

2 years ago