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notsponge [240]

3 years ago

13

If (3x2 + 22x + 7) ÷(x + 7) = 3x + 1, then (x + 7)( ) = . The check of the polynomial division problem shows that the product of

two polynomials is a polynomial. This supports the fact that the property is satisfied for polynomial multiplication.

1 answer:

azamat3 years ago

8 0

Answer:

$(x + 7) * (3x + 1) = (3x^2 + 22x + 7)$

Step-by-step explanation:

Given

$(3x^2 + 22x + 7) \div (x + 7) = 3x + 1$

$(x + 7) * (\ ) = [\ ]$

Required

Complete the blanks

We have:

$(3x^2 + 22x + 7) \div (x + 7) = 3x + 1$

Rewrite as:

$\frac{(3x^2 + 22x + 7) }{ (x + 7)} = 3x + 1$

Cross multiply

$(x + 7) * (3x + 1) = (3x^2 + 22x + 7)$

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Someone plz help!!!!!!

aliina [53]

Answer: X = 7√2

Step-by-step explanation:

Let first Consider triangle BDC,

Cos C = adjacent/ hypothenus

Cos C = 7 / x ...... (1)

Also, let consider triangle ABC

Cos C = adjacent / hypothenus

Cos C = x / 14 ....... (2)

Since angle C is the same, equate equation 1 to 2

7/ x = x / 14

Cross multiply

X^2 = 98

Make x the subject of formula

X = sqrt (98)

X = sqrt ( 49 × 2 )

X = sqrt (49) × sqrt (2)

X = 7 sqrt(2)

X = 7√2

8 0

2 years ago

8.7 × 10 to the 5th power in standard form

Andrej [43]

Answer would be 870000

6 0

3 years ago

Read 2 more answers

A box of mashed paragraphs flakes has a volume 220in. The box is 8 inches long and 11 inches tall. What is the width of the box

Taya2010 [7]

The width is 2.5 in. Volume = Length x Width x Height

$220 = 8 \times 11 \times x$

$\frac{220}{8} = x = 2.5$

5 0

3 years ago

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

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

or

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

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 $x\in A\cap B$ .

Observe the image below.

7 0

3 years ago

How can I simplify this and show work?

svp [43]

$\frac{ \sqrt{3}}{ \sqrt{5} - \sqrt{6} } \\ = \frac{ \sqrt{3}( \sqrt{5} + \sqrt{6}) }{( \sqrt{5} - \sqrt{6})( \sqrt{5} + \sqrt{6})} \\ = \frac{ \sqrt{15}+3 \sqrt{2} }{5-6} \\ =- \sqrt{15}-3 \sqrt{2}$
Please vote me the Brainliest one. Thank you <3

6 0

3 years ago