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3 years ago

Help as soon as possible pleaseee

Mathematics

2 answers:

mestny [16]3 years ago

8 0

Factor by grouping so your answer you be #1 (3x +2)(x-4)

KiRa [710]3 years ago

3 0

Answer:

(3x + 2)(x-4)

Step-by-step explanation:

3x² - 10x - 8 = 0

(splitting the middle term)

3x² -12x + 2x -8 = 0

3x(x-4) + 2(x-4) = 0

(3x + 2)(x-4) = 0

(3x + 2)(x-4) is the correct answer (option a)

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PLEASE HELP!! this is ratios

DIA [1.3K]

Answer:

1. 8:5

2. 17;4

3. 23:6

4. I don't know

6 0

3 years ago

Read 2 more answers

The longer leg of a right triangle is 4cm longer than the shorter leg. the hypotenuse is 8cm longer than the shorter leg. find t

Alexeev081 [22]

Longer leg = x + 4

Shorter leg = x

Hypotenuse = x + 8

Using Pythagorean Theorem:

a^2 + b^2 = c^2

(x + 4)^2 + x^2 = (x + 8)^2

x^2 + 8x + 16 + x^2 = x^2 + 16x + 64

2x^2 + 8x + 16 = x^2 + 16x + 64

2x^2 +8x = x^2 + 16x + 48

2x^2 - 8x = x^2 + 48

x^2 - 8x = 48

x^2 - 8x - 48 = 0

You can complete the square from here or use the quadratic formula.

Completing the square:

x^2 - 8x = 48

x^2 - 8x + (-8/2)^2 = 48 + (-8/2)^2

x^2 - 8x + 16 = 48 + 16

(x - 4)(x - 4) = 64 or (x - 4)^2 = 64

x - 4 = +√64 OR x - 4 = -√64

x - 4 = +8 OR x - 4 = -8

x = 12 OR x = -4

However, you can't use negative 4 as a length because your length needs to be a positive.

So x will be 12.

Shorter leg: 12

Longer leg: 12 + 4

Hypotenuse: 12 + 8

8 0

3 years ago

The equation shown below represents a circle. Which statement describes the key features of the circle that can be determined fr

bija089 [108]

Answer:

The circle has a center at (5.-1) and radius of 2 units.

Step-by-step explanation:

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2 years ago

(with steps please) Find the inverse Laplace transform, f(t), of the function: 16/(s-4)^3

Stells [14]

Answer: The required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

Step-by-step explanation: We are given to find the inverse Laplace transform, f(t), of the following function :

$F(s)=\dfrac{16}{(s-4)^3}.$

We have the following Laplace formula :

$L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}}\\\\\\\Rightarrow L^{-1}\{\dfrac{1}{(s-a)^{n+1}}\}=\dfrac{t^ne^{at}}{n!}.$

Therefore, we get

$f(t)\\\\=L^{-1}\{\dfrac{16}{(s-4)^3}\}\\\\\\=16\times\dfrac{t^{3-1e^{4t}}}{(3-1)!}\\\\\\=\dfrac{16}{2}t^2e^{4t}\\\\\\=8t^2e^{4t}.$

Thus, the required inverse transform of the given function is

$f(t)=8t^2e^{4t}.$

6 0

3 years ago

The equation a<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class=

andrezito [222]

Answer:

$\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)$

Step-by-step explanation:

We are given the equation:

$ax^2+bx+c=0$

Which has roots α and β.

And we want to express (α + 1)(β + 1) in terms of a, b, and c.

From the quadratic formula, we know that the two solutions to our equation are:

$\displaystyle x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}\text{ and } x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Let x₁ = α and x₂ = β. Substitute:

$\displaystyle \left(\frac{-b+\sqrt{b^2-4ac}}{2a} + 1\right) \left(\frac{-b-\sqrt{b^2-4ac}}{2a}+1\right)$

Combine fractions:

$\displaystyle =\left(\frac{-b+2a+\sqrt{b^2-4ac}}{2a} \right) \left(\frac{-b+2a-\sqrt{b^2-4ac}}{2a}\right)$

Rewrite:

$\displaystyle = \frac{\left(-b+2a+\sqrt{b^2-4ac}\right)\left(-b+2a-\sqrt{b^2-4ac}\right)}{(2a)(2a)}$

Multiply and group:

$\displaystyle = \frac{((-b+2a)+\sqrt{b^2-4ac})((-b+2a)-\sqrt{b^2-4ac})}{4a^2}$

Difference of two squares:

$\displaystyle = \frac{\overbrace{(-b+2a)^2 - (\sqrt{b^2-4ac})^2}^{(x+y)(x-y)=x^2-y^2}}{4a^2}$

Expand and simplify:

$\displaystyle = \frac{(b^2-4ab+4a^2)-(b^2-4ac)}{4a^2}$

Distribute:

$\displaystyle = \frac{(b^2-4ab+4a^2)+(-b^2+4ac)}{4a^2}$

Cancel like terms:

$\displaystyle = \frac{4a^2+4ac-4ab}{4a^2}$

Factor:

$\displaystyle =\frac{4a(a+c-b)}{4a(a)}$

Cancel. Hence:

$\displaystyle = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)$

Therefore:

$\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}$

4 0

3 years ago