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3 years ago

Solve the equation.

Mathematics

1 answer:

mylen [45]3 years ago

7 0

-9/15y+3/21=5/15y-14/21

Move 5/15y to the other side. Sign changes from +5/15y to -5/15y

-9/15y-5/15y+3/21=5/15y-5/15y-14/21

-14/15y+3/21=-14/21

Move 3/21 to the other side. Sign changes from +3/21 to -3/21.

-14/15y+3/21-3/21=-14/21-3/21

-14/15y=-14/21-3/21

-14/21-3/21=-17/21

-14/15y=-17/21

Multiply both sides by -15/14

-14/15y(-15/14)

Cross out 15 and 15, divide by 15 then becomes 1

Cross out 14 and 14, divide by 14 then becomes 1

1*1*y=y

-17/21*-15/14

Cross out 15 and 21 , divide by 3. 15/3=2, 21/3=7

17/7*5/14=85/98

Answer: c. y=85/98

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Pls help me I am freaking out. :(

kkurt [141]

Andre : Anything less than -100, for example -150

-150<-100

|-150|=150

Han: Anything between -100 & 100, for example 50

50> 100

|50|= 50

Lin: 100 or -100,

100> -100

|100|=100

-100=-100

|-100|=100

8 0

3 years ago

Below are two parallel lines with a third line intersecting them.

Scrat [10]

Answer:

x = 53 degree

Step-by-step explanation:

53 degree angle = x angle( alternate exterior angle)

Therfore, x = 53 degree

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3 0

2 years ago

Solve for a.

egoroff_w [7]

Answer: $a=\sqrt{\frac{30}{23} }$

Step-by-step explanation:

$23a^2=30$

Divide by 23

$a^2=\frac{30}{23}$

Extract the square root.

$a=\sqrt{\frac{30}{23} }$

3 0

3 years ago

One less than one fourth the square of a number.

Free_Kalibri [48]

Answer:

1/4m²-1

let m be the number.

8 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

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