What is the solution set of the quadratic inequality x^2+x-2>=0?

s344n2d4d5 [400]

The average salary is $18,725

To find the answer add up each job's salary then divide by the number of jobs.

28,700 + 18,600 + 14,66 + 13,00 = 74,900

Then divide 74,900 by 4 ( added 4 numbers )

6 0

3 years ago

FASTTTTTTTTT PPPPPPPPPPPPPPLLLLLLLLLLLLLLLLZZZZZZZZZZZZZZZ

Deffense [45]

Answer:

(a) B

(b) $2

Step-by-step explanation:

(a) Let's say the cost of a ticket is t and the cost of popcorn is p. Then we can write the two equations from the table:

12t + 8p = 184

9t + 6p = 138

We need to solve this, so let's use elimination. Multiply the first equation by 3 and the second equation by 4:

3 * (12t + 8p = 184)

4 * (9t + 6p = 138)

We get:

36t + 24p = 552

Subtract the second from the first:

36t + 24p = 552

- 36t + 24p = 552

________________

0 = 0

Since we get down to 0 = 0, which is always true, we know that we cannot determine the cost of each ticket because there is more than one solution (infinitely many, actually). The answer is B.

(b) Our equation from this, if we still use t and p, is:

5t + 4p = 82

Now, just choose any of the two equations from above. Let's just pick 9t + 6p = 138. Now, we have the system:

5t + 4p = 82

9t + 6p = 138

To solve, let's use elimination again. Multiply the first equation by 6 and the second one by 4:

6 * (5t + 4p = 82)

4 * (9t + 6p = 138)

We get:

30t + 24p = 492

36t + 24p = 552

Subtract the second from the first:

36t + 24p = 552

- 30t + 24p = 492

________________

6t + 0p = 60

So, t = 60/6 = $10. Plug this back into any of the equations to solve for p:

5t + 4p = 82

5 * 10 + 4p = 82

50 + 4p = 82

4p = 32

p = 32/4 = $8

So the ticket costs 10 - 8 = $2 more dollars than the popcorn.

5 0

3 years ago

Read 2 more answers

) What is the surface area? 10 yd 8 yd 10 yd 12 yd 10 yd

Anastaziya [24]

This is a formula chart to help(i would’ve given you the answer but you didn’t give the sharp)

6 0

3 years ago

How many symmetry does maple leaf have

Travka [436]

Only one Line.........

6 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