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4 years ago

Can I have help with 43 and 44 I need to see how to do them thanks.

Mathematics

2 answers:

steposvetlana [31]4 years ago

7 0

1. 3x^5 -75x³

=3x³(x²-25)

=3x³(x²-5²)

=3x³(x-5)(x+5)

2. 81c²+72c+16

=81c²+36c+36c+16

=9c(9c+4)+4(9c+4)

=(9c+4)(9c+4)

=(9c+4)²

aliina [53]4 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

(43)

3 $x^{5}$ - 75x³ ← factor out 3x³ from each term

= 3x³(x² - 25) ← this is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , thus

x² - 25 = x² - 5² = (x - 5)(x + 5)

Thus

3 $x^{5}$ - 75x³ = 3x³(x - 5)(x + 5)

(44)

81c² + 72c + 16 ← is a perfect square of the form

(ac + b)² = a²c² + 2abc + b²

Compare coefficients of like terms

a² = 81 ⇒ a = $\sqrt{81}$ = 9

b² = 16 ⇒ b = $\sqrt{16}$ = 4

and 2ab = 2 × 9 × 4 = 72

Thus

81c² + 72c + 16 = (9c + 4)²

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Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .