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3 years ago

PLEASE HELP ME SO I CAN MOVE ON FROM THIS TOPIC

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

4 0

Answer:

3rd option

Step-by-step explanation:

( $\frac{f}{g}$ )(x)

= $\frac{f(x)}{g(x)}$

= $\frac{2x^2-5x-3}{2x^2+5x+2}$ ← factorise numerator and denominator

= $\frac{(x-3)(2x+1)}{(x+2)(2x+1)}$ ← cancel (2x + 1) on numerator/ denominator

= $\frac{x-3}{x+2}$

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Solve this system of equations using elimination. X-6=-17 -2x+12y=7

AfilCa [17]

Answer:

Add the equations in order to solve for the first variable. Plug this value into the other equations in order to solve for the remaining variables.

Point Form:

(−11,−5/4)

Equation Form:

x=−11,y=−5/4

8 0

2 years ago

The discriminant of a quadratic equation is equal to -8, which statement describes the roots?

Maslowich

Answer:

(a) There are two complex roots

Step-by-step explanation:

The discriminant of a quadratic function describes the nature of its roots:

negative: two complex roots
zero: one real root (multiplicity 2)
positive: two distinct real roots.

Your discriminant of -8 is negative, so it indicates ...

There are two complex roots

_____

Additional comment

We generally study polynomials with real coefficients. These will never have an odd number of complex roots. Their complex roots always come in conjugate pairs.

6 0

2 years ago

One fifth of a number n is equal -7 write equation and true false or open sentence

Otrada [13]

One fifth of a number n is equal -7 as an equation would look like this:

1/5 n = -7 / * 5 (both sides)
n = -35

So if number n equals -35, then it's true that 1/5n equals -7.

4 0

3 years ago

Read 2 more answers

4/6 is equivalent to what fraction?

rusak2 [61]

It is equivalent to 4/6 so yeah yay.

6 0

3 years ago

Read 2 more answers

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

2 years ago