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Olenka [21]
3 years ago
11

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. <br> X-6=-17<br> -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:

  • <u>negative</u>: two complex roots
  • <u>zero</u>: one real root (multiplicity 2)
  • <u>positive</u>: two distinct real roots.

__

Your discriminant of -8 is <em>negative</em>, so it indicates ...

  There are two complex roots

_____

<em>Additional comment</em>

We generally study polynomials with <em>real coefficients</em>. 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 <br> 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
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