6x4? multiplication answer

If a quadratic equation with real coefficients has a discriminant of 3 then the two roots must be

lukranit [14]

In a quadratic equation

q(x) = ax^2 + bx + c

The discriminant is = b^2 - 4ac

We have that discriminant = 3

If b^2 - 4ac > 0, then the roots are real.

If b^2 - 4ac < 0 then the roots are imaginary

<span>In this problem b^2 - 4ac > 0 3 > 0 </span>

then the two roots must be real

6 0

3 years ago

Please assist me in my time of need.

vaieri [72.5K]

Answer:

I shall assist thee.

Step-by-step explanation:

Being me, I would say the answer is A. or B. But i'd prolly go with A tbh.

And oofers on me if this ain't what you need.

4 0

3 years ago

Omar works as a tutor for $15 an hour and as a waiter for $7 an hour. This month, he worked a combined total of 83 hours at his

Alborosie

Answer: $8t+581$

Step-by-step explanation:

<h3> The complete exercise is: "Omar works as a tutor for $15 an hour and as a waiter for $7 an hour. This month, he worked a combined total of 83 hours at his two jobs. Let "t" be the number of hours Omar worked as a tutor this month. Write an expression for the combined total dollar amount he earned this month."</h3><h3 />

Let be "t" the number of hours Omar worked as a tutor this month and "w" the number of hours Omar worked as a waiter this month.

Based on the data given in the exercise, you know that Omar worked a combined total of 83 hours this month.

Then, you can represent the number of hours he worked as a waiter this month with this equation:

$w=83-t$

Since he earns $15 per hour working has a tutor and $7 per hour working as a waiter, you can write the following expresion to represent the total money earned:

$15t+7w$

Since $w=83-t$ , you can substitute it into the expression and then simplify it in order to find the final expression that represents the total amount of money Omar earned this month.

This is:

$15t+7(83-t)=15t+581-7t=8t+581$

4 0

4 years ago

I need help , I don’t understand this

marta [7]

#2. First, we factor each polynomial. Then, if any terms on both the top and the bottom of the fraction match, they cancel out. So... we do just that. You end up with:

$\frac{x(x-4)}{(x+9)(x-4)}$

Notice there's an (x-4) on both top and bottom. So they cancel out. That leaves us with your answer of $\frac{x}{(x+9)}$

#3. We do the same thing as above then multiply and simplify. In the interest of space, I'll cut straight to some simplification.

$\frac{2(x+2)^{3} }{6x(x+2)} ( \frac{5}{(x-2)^{2} } )$

Now we start cancelling. For the first fraction, there are 3 (x+2)'s on top and 1 on the bottom so we will cancel out the one on the bottom and leave 2 (x+2)'s on top. There are no more polynomials to cancel out so now we multiply across:

$\frac{10(x+2)^{2} }{6x(x-2)^{2} }$

10 and 6 share a GCF of 2 so we divide both of those by 2. This leaves us with the final answer of:

$\frac{5(x+2)^{2} }{3x(x-2)^{2} }$

#4. This equation introduces division and because of it, we must flip the second fraction to make the division sign into a multiplication symbol. Again for space, I'll flip the fraction and simplify in one step.

$\frac{3(x+2)(x-2)}{(x+4)(x-2)} ( \frac{x+4}{6(x+3)})$

Now we do our cancelling. First fraction has (x - 2) in the top and bottom. They're gone. The first fraction has a (x + 4) on the bottom and the second fraction has one on the top. Those will also cancel. This leaves you with:

$\frac{3(x+2)}{6(x+3)}$

3 and 6 share a GCF of 3 so we divide both numbers by this. This leaves you with your final answer:

$\frac{x+2}{2(x+3)}$

#5. We are adding so we first factor both fractions and see what we need to multiply by to make the denominators the same. I'll do the former first. (10 - x) and (x - 10) are not the same so we multiply the first equation (top and bottom) by (x - 10) and the second equation by (10 - x). Because they will now have the same denominator we can combine them already. This gives us:

$\frac{(3+2x)(x-10)+(13+x)(10-x)}{(10-x)(x-10)}$

Now we FOIL each to expand and then simplify by combining like terms. Again for space, I'm just showing the result of this; you end up with:

$\frac{x^{2}-20x+100}{(10-x)(x-10)}$

Now we factor the top. This gives you 2 (x - 10)'s on top and one on bottom. So we just leave one on the top and cancel the bottom one out. This leaves you with your answer:

$\frac{x+10}{10-x}$

#6. Same process for this one so I won't repeat. I'll just show the work.

$\frac{3}{(x-3)(x+2)} + \frac{2}{(x-3)(x-2)}$ becomes

$\frac{3(x-2) + 2(x+2)}{(x-3)(x+2)(x-2)}$ which equals

$\frac{3x - 6 + 2x + 4}{(x-3)(x+2)(x-2)}$ giving you the final answer

$\frac{5x - 2}{(x-3)(x+2)(x-2)}$

#7. For this question we find the least common denominator to make the denominators match. For 5, x, and 2x, the LCD is 10x. So we multiply top and bottom of each fraction by what would make the bottom equal 10x. This rewrites the fraction as:

$\frac{3x}{5} ( \frac{2x}{2x}) * ( \frac{5}{x}( \frac{10}{10}) - \frac{5}{2x} ( \frac{5}{5}))$

Simplify to get:

$\frac{3x}{5} * ( \frac{25}{10x})$

After simplifying again, you end up with your final answer:

$\frac{3}{2}$

8 0

3 years ago

Determine if AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯ are parallel, perpendicular, or neither. Given: A (−1, 3), B (0, 5), C (2, 1), D (6, −1)

dolphi86 [110]

Answer:

perpendicular

Step-by-step explanation:

To determine if AB and CD are parallel, perpendicular, or neither, we need to get the slope of AB and CD first

Given A (−1, 3), B (0, 5),

Slope Mab = 5-3/0-(-1)

Mab = 2/1

Mab = 2

Slope of AB is 2

Given C (2, 1), D (6, −1)

Slope Mcd = -1-1/6-2

Mcd = -2/4

Mcd = -1/2

Slope of CD is -1/2

Take their product

Mab * Mcd = 2 * -1/2

Mab * Mcd = -1

Since the product of their slope is -1, hence AB and CD are perpendicular

5 0

3 years ago