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3 years ago

CAN SOMEONE HELP ME PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

Mathematics

1 answer:

Anettt [7]3 years ago

7 0

The LCD is the product of (x -2)(x + 1)
To clear the fractions, multiply each term by the part of the LCD that the denominator is missing.
x(x + 1) + (x - 2)(x - 1) = -1(x - 2)(x + 1)
Distribute and foil, then combine like terms.
x^2 + x + x^2 - 3x + 2 = -x^2 + x + 2
2x^2 - 2x + 2 = -x^2 + x + 2
add x^2 to both sides
3x^2 -2x + 2 = x + 2
subtract from both sides
3x^2 - 3x + 2 = 2
subtract 2 from both sides.
3x^2 - 3x = 0
Factor out the GCF
3x(x - 1) = 0
Set each factor equal to zero and solve.
3x = 0 x - 1 = 0
x = 0 x = 1

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For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th

timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

$(a,b)=\left \{ x\in\Re : a$

2. Functions

1) $\Re \rightarrow \Re \\ f(x)=x$

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2) $\Re\geq0\rightarrow\Re , f(x)=x+1\\$

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3) $f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)$

The cosine function is not injective, bijective nor surjective.

4) $f:\Re\rightarrow\Re \:f(x)=ex$

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5) Quite unclear format

$6) \:f:\Re\rightarrow(0,\infty), f(x) =ex$

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

$7) f:\Re\geq 0 \rightarrow \Re\geq0, f(x) =x^{4}$

Not injective nor surjective therefore not bijective too.

$8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}$

$f(-1)=1, f(2)=4, f(-3)=9$

Injective (one to one), Surjective, and Bijective.

$10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.$

Surjective.

$11.f:R\geq 0[-1,1], f(x) = 0\\$

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

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3 years ago

Given h(x)=3x+4, solve for x when h(x)=1

aivan3 [116]

Answer:

Step-by-step explanation:

Given,

h ( x ) = 1

To find : -

h ( x ) = 3x + 4

h ( 1 ) = 3 ( 1 ) + 4

= 3 + 4

h ( 1 ) = 7

3 0

3 years ago

Read 2 more answers

Are all terminating and repeating decimals rational numbers?

olganol [36]

Answer:

Yes they are rational numbers

Step-by-step explanation:

Rational numbers are Any number that can be written in fraction form is a rational number. This includes integers, terminating decimals, and repeating decimals as well as fractions. ... So, any terminating decimal is a rational number.

4 0

3 years ago

H. Goes through (-3, -2) and parallel to y = 5 x +5

olchik [2.2K]

Answer: y = 5x + 13

Step-by-step explanation:

Parallel means it has the same slope.

Given line y = 5x + 5 --> the slope is 5, so parallel slope is 5

Use the Point-Slope formula: y - y₁ = m(x - x₁) such that

m = 5
(x₁, y₁) = (-3, -2)

y + 2 = 5(x + 3) input m and (x₁, y₁) into the Point-Slope formula

y + 2 = 5x + 15 distributed

y = 5x + 13 subtracted 2 from both sides

7 0

3 years ago

I will get my points but answer it correctly and no LINKS!!! (factotise by using a suitable identity X5-X)

9966 [12]

x(x-1)(x+1)(x²+1)=0

Hope it helps you

6 0

3 years ago