All categories

4 years ago

How do you distribute a negative symbols

Mathematics

1 answer:

Sindrei [870]4 years ago

8 0

Answer:

-5.90 = y

Step-by-step explanation:

12(5 +2y) = 4y -( 5 - 9y)

1. Distribute

60 + 24y = 4y -5 + 9y

2. Add like terms

60 + 24y = -5 + 13y

3. Add 5 on both sides

60 + 24y +5 = -5 + 13y +5

65 + 24y = 13y

4. Subtract 24y on both sides

65 + 24y -24y = 13y -24y

65 = -11y

5. Divide -11 on both sides

65/-11 = -11y/-11

-5.90 = y

You might be interested in

Kamil and Sean share some money in the ratio 3:5. Sean gets $56 more than Kamil. How much do they get each?

liq [111]

Let Kamil get x. Therefore Sean would get (x + 56).

Kamil and Sean are in the ratio: 3:5

That means: x : (x + 56) = 3 : 5

x / (x + 56) = 3/5

5*x = 3*(x + 56)

5x = 3*x + 3*56

5x = 3x + 168

5x - 3x = 168

2x = 168

x = 168 / 2

x = 84

Therefore Kamil had, x = 84, and Sean had (x + 56) = 84 + 56 = 140

Kamil had $84 and Sean had $140

3 0

3 years ago

Use De Moivre’s Theorem to compute the following: <img src="https://tex.z-dn.net/?f=%5B3%28cos27%2Bisin27%29%7D%5E5" id="TexForm

zhannawk [14.2K]

Hello:
Use De Moivre’s Theorem :
(3(cos27 +isin27)^5 = 3^5( cos(27 × 5) +isin(27 × 5))
= 3^5 ( cos(135)+i sin(135))
= 3^5(-√2/2+i √2/2)
because : cos(135) = -√2/2 and sin(135) = √2/2
(3(cos27 +isin27)^5 = (- 3^5√2/2)+ i ( 3^5√2/2) ...(form : a+ib when
a= (- 3^5√2/2) and b = ( 3^5√2/2)

4 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

Determine the approximate value of θ. A. 38.9 B. 51.1 C. 37.9 D. 52.1

Pepsi [2]

Answer:

Option B. $51.1^o$

Step-by-step explanation:

we know that

In the right triangle of the figure

The sine of angle theta is equal to divide the opposite side to angle theta by the hypotenuse

$sin(\theta)=\frac{DF}{DE}$

substitute the given values

$sin(\theta)=\frac{7}{9}$

using a calculator

$\theta=sin^{-1}(\frac{7}{9})=51.1^o$

8 0

4 years ago

If logb 5=0.41, find logb 12

Hitman42 [59]

Determine which one of the following is the correct application of the change of base formula?

a)log3 14= log10 14/3
b)lo4 17= log10 4/ log10 17
c)log8 3= log10 3/log10 8
d)log9 32 = log10 32-log10 9

The change of base formula is as follows:
log to the base 10 of x
log to the base b of x = ----------------------------------
log to the base 10 of b

Compare this to answer choice C:

c)log8 3= log10 3/log10 8 Here x = 3 and b = 8. This is the only correct choice.

6 0

3 years ago