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2 years ago

The diagonals of a rhombus are perpendicular.

Mathematics

1 answer:

Ivahew [28]2 years ago

3 0

The rhombus always has perpendicular diagonals.

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4/5 of a number is 16. What is the number?

Zepler [3.9K]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Sierra is saving money to buy a bike the bike costs $245. she has $125 saved and each week she adds $15 to her savings. how many

beks73 [17]

You should start with an equation.
$245 = 125 + 15x$
The 'x' represents the week.

so now, we have to subtract 125 from the total amount which is 245.
$245 - 125 = 120$
now we have
$120 = 15x$
to find 'x', we must divide 15 on each side
$\frac{120}{15} = \frac{15x}{15}$
we are now left with
$8 = x$
that means Sierra will have 8 weeks until she has enough money to buy the bike

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2 years ago

Cody's Gym charges a $15 fee to join and $30 per month. Tyler's Gym charges a $30 fee to join and $20 per month. Write an inequa

posledela

Answer:

After (3/2) months Tyler's gym would be a better deal.

Step-by-step explanation:

Let x the number of months

Cody's Gym charges a $15 fee to join and $30 per month

y₁= 15 + 30x (1)

Tyler's Gym charges a $30 fee to join and $20 per month

y₂ = 15 + 30x (2)

See the attached figure which represent the graph of y₁ (with blue color) and y₂ (with red color)

The point of intersection between y₁ and y₂ is ( 3/2 , 60)

After 3/2 months Tyler's Gym charges will be less than Cody's Gym charges

So, the inequality will be:

After (3/2) months Tyler's gym would be a better deal.

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2 years ago

A. Sin(0)=0.6 B.Cos(0)=0.75 C.Tan(0)=1.3 D. Cos(0)=0.8

LuckyWell [14K]

Answer:

its e

Step-by-step explanation:

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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

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