All categories

3 years ago

What is the product of (4x + 6 ) and (2x - 5)

Mathematics

1 answer:

Mademuasel [1]3 years ago

4 0

(4x+6)(2x-5)

= 8x^2 - 20x + 12x - 30

= 8x^2 - 8x - 30

You might be interested in

Two neighbors, wilma and betty, each have a swimming pool. both wilma's and betty's pools hold 9900 gallons of water. if wilma's

s2008m [1.1K]

To solve this problem we will start by calculating time needed for each of them to fill the pool.

We have formula:
Volume = rate * time
Or
time = volume / rate

Wilma:
time = 9900 / 900
time = 11h

Betty:
time = 9900 / 500
time = 19.8h

Now we substract these two numbers:
time_difference = 19.8 - 11 = 8.8h

Betty needs 8.8 hours more than Wilma to fill the pool.

3 0

3 years ago

What is the value of k in y = kx, if y = 12 when x = 24?

Brrunno [24]

Answer:

k = 0.5

Step-by-step explanation:

Substitue y = 12, x = 24 into y = kx.

12 = k × 24

24k = 12

k = 12 ÷ 24

k = 0.5 (final answer)

6 0

3 years ago

Read 2 more answers

Needs to be over P_4(C), I know how to do over P_4(R).

nasty-shy [4]

Answer:

defendant who, before being served with process, timely returns a waiver need not serve an answer to the complaint until 60 days after the request was sent—or ...

Step-by-step explanation:

7 0

2 years ago

Factor the polynomial, x2 + 5x + 6

patriot [66]

Answer:

Choice b.

$x^{2} + 5\, x + 6 = (x + 3)\, (x + 2)$ .

Step-by-step explanation:

The highest power of the variable $x$ in this polynomial is $2$ . In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to $0$ .)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

Apply the quadratic formula to find the two roots that would set this quadratic polynomial to $0$ . The discriminant of this polynomial is $(5^{2} - 4 \times 1 \times 6) = 1$ .

$\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}$ .

Similarly:

$\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}$ .

By the Factor Theorem, if $x = x_{0}$ is a root of a polynomial, then $(x - x_0)$ would be a factor of that polynomial. Note the minus sign between $x$ and $x_{0}$ .

The root $x = -2$ corresponds to the factor $(x - (-2))$ , which simplifies to $(x + 2)$ .
The root $x = -3$ corresponds to the factor $(x - (-3))$ , which simplifies to $(x + 3)$ .