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4 years ago

2x+4y=1 3x-5y=7 solve the simultaneous equations

Mathematics

1 answer:

WITCHER [35]4 years ago

4 0

Answer:

y= -1/2

x=3/2

Step-by-step explanation:

first isolate x

2x+4y=1

2x=1-4y

x=(1-4y)/2

subsitute it in the second equation

look at the pic

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The price of an item has increased 15% since last year. However, a person can buy the item for a 25% employee discount. The empl

ANEK [815]

Answer: $200

Explanation:

The employee pays $172.50 and he got 25% employee discount.

If the cost is $100, he pays $(100 - 25) = $75.

So, employee pays =

100

⋅

172.50

= $230.00.

Again, price was increased by 15% last year. So whose cost is $100

is sold by $(100+15) = $115.

When sell price is $115, then cost price is $100. Then,

when sell price is $230, then cost price is

100

115

⋅

230

= $200.

5 0

3 years ago

Plutonium-240 decays according to the function L where o

PIT_PIT [208]

Answer:

$\boxed{\textbf{1700 yr}}$

Step-by-step explanation:

$-\dfrac{\text{d}L}{\text{d}t} = kL\\\\\dfrac{dL}{L}=-kdt\\\\\ln L = -kt + C\\\text{At t = 0, L = L$_{0}$, so C = $\ln L_{0}$}\\\ln L = -kt + \ln L_{0}\\\ln L_{0} - \ln L = kt\\\\\ln \dfrac{L_{0}}{L} =kt$

Data:

L₀ = 24 g

L = 20 g

k = 0.000 11 yr⁻¹

Calculation:

$\ln \dfrac{24}{20} =0.000 11t\\\\\ln 1.2 = 0.000 11t\\\\0.1823 = 0.000 11t\\\\t = \dfrac{0.1823}{0.000 11} = \textbf{1700 yr}\\\\\text{It will take } \boxed{\textbf{1700 yr}}\text{ for the polonium to decay to 20 g}$

8 0

3 years ago

8 w + 12 = -12<br> Helpppppp

aliina [53]

Answer:

w = -3

Step-by-step explanation:

8w + 12 = -12

use inverse operations:

8w + 12 = -12

-12 -12

8w = -24

/8 /8

w = -3

7 0

3 years ago

Read 2 more answers

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

A company sells widgets. The amount of profit, y, made by the company, is related to the selling price of each widget, x, by the

RUDIKE [14]

9514 1404 393

Answer:

$1571

Step-by-step explanation:

If we assume that y is profit in dollars, the maximum value of y is revealed by a graph to be ...

ymax = $1571

The maximum profit the company can make is $1571 at a selling price of $45 each.

_____

The function can be written in vertex form as ...

y = -(x^2 -90x) -454

y = -(x^2 -90x +45^2) -454 +45^2

y = -(x -45)^2 +1571

Because the leading coefficient is negative, we know this vertex represents a maximum. (x, y) = (45, 1571) at the vertex. Maximum profit is $1571.

8 0

3 years ago