Could somebody please help me

Can anyone solve this for me

emmainna [20.7K]

Answer:

(-2, -4.5)

Step-by-step explanation:

We can solve this equation with substitution.

x=2y+7

3x-2y=3

We can "substitute" 2y+7 for x into the second equation:

3(2y+7)-2y=3

Distribute the 3

6y+21-2y=3

Add like terms

4y+21=3

Subtract 21 from both sides

4y=-18

Divide both sides by 4 to isolate y

y=-4.5

Plug -4.5 back in for y:

x=2(-4.5)+7

x=-9+7

x=-2

(x,y)=(-2,-4.5)

7 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

)The equation of the line of best fit of a scatter plot is y = −4x − 6. What is the the y-intercept?

Gekata [30.6K]

Well the y-intercept is where a line crosses the y-axis
y=-4x-6
think of it as y=mx+b
m is always going to be your slope
b is always going to be your y-intercept
y-intercept of y=-4x-6
is -6

8 0

4 years ago

Read 2 more answers

I help me please I need these answers

Ivan

Answer:

G i think that is the answer

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

the length of your garden must exceed the width of the garden by 3 feet. you only have 58 feet of fencing to go around the perim

DedPeter [7]

First let us represent
x-width
x+3-length

Since it has a width and length, we can say it is a rectangle so the formula for the perimeter of it is 2(l+w)

p=2(l+w)
58=2(x+3+x)
58=(2x+3)2
58=4x+6
52=4x
x=13

the width is 13 and the length is 16

7 0

3 years ago