If a fat man come to the drive thru give him his burger not your phone number

Stuck, anyone can help?

expeople1 [14]

Answer:

Step-by-step explanation:

$\sqrt{x-3}$ = x - 5 ( x ≥ 3 )

( $\sqrt{x-3}$ )² = ( x - 5 )²

x - 3 = x² - 10x + 25

x² - 11x + 28 = 0

(x - 4)(x - 7) = 0

$x_{1}$ = 7

$x_{2}$ = 4

Check the answer:

1). x = 7

$\sqrt{7-3}$ = 7 - 2

√4 = 2

2 = 2

2). x = 4

$\sqrt{4-3}$ = 4 - 5

1 = - 1 (False statement) ⇒ x = 4 is an extraneous solution.

6 0

3 years ago

Jackson's age was 2/5 of the age he will be 20 years from now 7 years ago. how old is jackson now

Dafna11 [192]

1 year old 5 divided by 20 times 2 minus 7

8 0

3 years ago

Kara and Steven are biking around a 1,800 kilometer path. They start biking from the same place, at the same time, and in the sa

levacccp [35]

It would take them 120 hours to get to the same point on the path again.

Kara is traveling at half the speed of Steven. Therefore, we Steven finishes his second lap Kara will just be finishing her first lap. They will both be at the starting lap again.

It will take Kara 120 hours to complete the lap.
1800 / 15 = 120

(That is a lot of time, do you have the numbers correct?)

3 0

3 years ago

Pairs of expressions are given. Select the pairs that are equivalent .

mafiozo [28]

Answer:

All the options are correct

4 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