All categories

3 years ago

Solve the equation: 3x - 2/4 = 2x – 8 Explain your reasoning / show work:

Mathematics

2 answers:

dezoksy [38]3 years ago

6 0

X=-15/2 is the answer

agasfer [191]3 years ago

4 0

X= -15/2
Reduce the fraction
Move terms
Collect like terms and calculate

You might be interested in

You are given a box of 100 silver dollars, all facing heads up. You are instructed to shake the box 2 times; after the second sh

Nimfa-mama [501]

100 divided by 2 is 50 and then divided by 6 is 8.333333 but you just round that to 8$.

8 0

3 years ago

Read 2 more answers

What is a construction (in geometry)

just olya [345]

Answer:

A. A construction is a figure drawn using only a straightedge and a compass

8 0

3 years ago

If a point is randomly selected from the rectangular area of the graph, what is the probability that it will be in the blue regi

nika2105 [10]

Answer:

Step-by-step explanation:

just took it

5 0

3 years ago

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

Dvinal [7]

For part (a), you have

$\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}$
$x=a(x-2)+b(x+3)$

If $x=2$ , then $2=b(2-3)\implies b=-2$ .

If $x=-3$ , then $-3=a(-3-2)\implies a=\dfrac35$ .

So,

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

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

In the remainder term, the denominator $x^2+x+2$ can't be factorized into linear components with real coefficients, since the discriminant is negative $(1-4\times1\times2=-7)$ . However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

$x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i$
$\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)$
$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
$x+2=a\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)+b\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)$

When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
$\dfrac32-\dfrac{\sqrt7}2i=-\sqrt7ib$
$b=\dfrac12+\dfrac3{2\sqrt7}i=\dfrac1{14}(7+3\sqrt7i)$

When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
$\dfrac32+\dfrac{\sqrt7}2i=\sqrt7ia$
$a=\dfrac12-\dfrac3{2\sqrt7}i=\dfrac1{14}(7-3\sqrt7i)$

So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

5 0

3 years ago

Read 2 more answers

If sin(x) =-1/2 and tan(x) is negative what is cos(2x)

Keith_Richards [23]

Answer:

positive

Step-by-step explanation:

Using the acronym ASTC (all students take calculus), this means that if sin and tan are negative, they must be in the fourth quadrant, which means that cos is positive.

ASTC is an acronym to help students remember which of the three basic trigonometric functions are positive in which quadrants.

A: all

S: sin

T: tan

C: cos

It starts from the first quadrant: all the sin, tan, cos are positive in the first quadrant, only sin is positive in the second quadrant, only tan is positive in the third quadrant, only cos is positive in the fourth quadrant.

3 0

3 years ago

Solve the equation: 3x - 2/4 = 2x – 8 Explain your reasoning / show work:​

Solve the equation: 3x - 2/4 = 2x – 8 Explain your reasoning / show work: