If I multiplied these binomials (x - 3) + (4x + 2) what would be the final product?

Mathematics

2 answers:

zlopas [31]3 years ago

8 0

Answer: 4x^2 - 10x - 6

Step-by-step explanation:

You can use FOIL to do this. First terms, Outer terms, Inner terms, Last terms.

(x - 3)(4x + 2)

First terms: x * 4x = 4x^2

Outer terms: x * 2 = 2x

Inner terms: -3 * 4x = -12x

Last terms: -3 * 2 = -6

Now add these together:

4x^2 + 2x - 12x - 6 = 4x^2 - 10x - 6

Lelechka [254]3 years ago

5 0

Answer:

4x^2 - 10x - 6

Step-by-step explanation:

If you truly want to multiply these binomials, remove the " + " sign and replace it with " * " for multiplication:

(x - 3) (4x + 2) = 4x^2 + 2x - 6 - 12x

Combining like terms, we get 4x^2 - 10x - 6

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Let x1 = 12, y1 = 15, and y2 = 3. Let y vary inversely with x. Find x2.

Ostrovityanka [42]

We have that an inverse variation in a function is given by:
$y = k / x
$
We must find the value of the constant k.
To do this, we substitute the values of x1 = 12, y1 = 15.
We have then:
$15 = k / 12
$
Clearing k we have:
$k = (15) * (12)

k = 180$
Substituting the value of k we have that the inverse variation is:
$y = 180 / x
$
For y2 = 3 we have:
$3 = 180 / x2
$
Clearing x2 we have:
$x2 = 180/3$
$x2 = 60$
Answer:
$x2 = 60$

3 0

4 years ago

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

nordsb [41]

$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7$