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2 years ago

Simple math! ( i'll mark you)

Mathematics

2 answers:

Studentka2010 [4]2 years ago

8 0

Answer:9,540

Step-by-step explanation:

LenKa [72]2 years ago

5 0

Answer:

234

Step-by-step explanation:

Glad to help mark me

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The price of a stock decreased $2 per day for four consecutive days. What was the total change in value of the stock over the fo

Elena L [17]

It changes by 2 for four days, so it's -$8

6 0

2 years ago

Read 2 more answers

State what additional information is required in order to know that the triangles are congruent by AAS. part 2

SVEN [57.7K]

Answer:

Step-by-step explanation:

The trick is to find the third angle

180 - A - <BCA = <CBA

180 - X - <BCX = <CBX

Everything else is OK

<CBA = <CBX Equals equated to an equal express = an equal expression.

Now by a little slight of hand, you get the two triangles to be equal by ASA, which always works.

Cheating you say? There is no such thing as cheating if it correct and it works.

8 0

2 years ago

Solve the equation 7/z-1 = 9/z+4

algol [13]

Answer:

z is equal to 18.5

Step-by-step explanation:

When an equation is in the format a/b = c/d, the quickest way to simplify it is to cross multiply. Each side is multiplied by the other sides denominator, eliminating the division:

$\frac{7}{z-1} = \frac{9}{z + 4}\\7(z + 4) = 9(z - 1)\\7z + 28 = 9z - 9\\7z - 9z = -9 - 28\\-2z = -37\\z = 37/2\\z = 18.5$

So z is equal to 18.5

3 0

2 years ago

80 marked up by 37.5 percent

Mariana [72]

Make 37.5% into decimal form : Since it is being marked up, add 100% = 1.375
1.375 x 80
1.375
x 80
_________
0000
+11000
_________
110.000

80 marked up by 37.5% is 110.

7 0

2 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

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