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3 years ago

Which steps could be part of the process in algebraically solving the system of equations, y + 5x = x2 + 10 and y = 4x –

Mathematics

1 answer:

mafiozo [28]3 years ago

3 0

Answer:

0 = x2 - 9x + 20

Step-by-step explanation:

y + 5x = x² + 10 Simplify this equation

- 5x - 5x

y = x² - 5x + 10

Set both equations equal to each other since they both equal y

x² - 5x + 10 = 4x - 10

-4x - 4x

x² - 9x + 10 = -10

+ 10 + 10

x² - 9x + 20 = 0

If this answer is correct, please make me Brainliest!

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Hope u can see this I will give 10 points to start of sorry for pushing u guys to type

Kruka [31]

Right triangles. There is a 90 degree angle on both

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3 years ago

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murzikaleks [220]

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2 years ago

7 qt. of a 27% saline solution was mixed

mamaluj [8]

Answer:

24% saline

Step-by-step explanation:

Find how much saline is in both solutions:

7(0.27)

= 1.89

3(0.17)

= 0.51

Add this together:

1.89 + 0.51

= 2.4

Since there are 10 quarts in total, the concentration of the mixture can be found by dividing 2.4 by 10:

2.4/10

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4 0

3 years ago

Normal form [see attachment]

musickatia [10]

I just did one of these. I guessed that normal form is

$x \cos \phi + y \sin \phi - p = 0$

Here' that's

$x (\sqrt{3}/2) + y (1/2) - 10 = 0$

$\sqrt{3} x + y - 20 = 0$

Choice b

4 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

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