Solve|3-v|<6

1.Write the inequality as two inequalities without absolute value.

2.Solve the inequality and write the solution set.

Answers

Answer 1

Answer: |3-v| < 6

1) Write the inequality as two inequalities without absolute value:
1st inequality: 3 < 6
2nd inequality: -v < 6

2)
|3-v| < 6
-3 -3
|-v| < 3
v < 3

v = 2

3 - v < 6
3 - 2 < 6
1 < 6

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Approximately how many miles are in a light-year?a. 6 × 106
b. 6 × 109
c. 6 × 1012
d. 6 × 1015

Answers

1 light year = 5.879 x 10^12
So approximately, the answer is C

Answer:

its C

Step-by-step explanation:

A 6 sided die was rolled 9 times. what is the mode of these rolls

Answers

the rolls are what.. ?

F(x) = 7x + 7, g(x) = 6x2

Find (fg)(x).

Answers

$f(x)=7x+7;\ g(x)=6x^2\n\n(fg)(x)=(7x+7)\cdot6x^2=42x^3+42x^2$

Ethan stands 256 cm at point C from a lamp post AB. He is 180 cm tall and his shadow from the lamp is 144 cm long. Find the height AB, in metres, of the lamp post.

Answers

Answer:500 cm

Step-by-step explanation:

Given

height of ethan = 180 cm

length of shadow =144 cm

Distance of ethan from lamp =256 cm

from diagram, we can write

$\Rightarrow tan \theta =(FC)/(EC)\quad \ldots(i)$

$\Rightarrow \tan \theta =(AB)/(BE)\quad \ldots (ii)$

From (i) and (ii) we get

$\tan \theta =(FC)/(EC)=(AB)/(BE)$

$\Rightarrow (180)/(144)=(h)/(144+256)$

$\Rightarrow h=(180)/(144)* 400$

$\Rightarrow h=500\ cm$

So, height of lamp post AB is $500\ cm$

Let a=x^2+4. Use a to find the solutions for the following equation: (x^2+4)^2+32=12x^2+48. Which one of the following are solutions for x? Select any/all that apply. -8, -2, 4, 0, 2, -4, 8

Answers

$(x^2+4)^2+32=12x^2+48 \n(x^2+4)^2+32=12(x^2+4) \ \ \ |-12(x^2+4) \n(x^2+4)^2-12(x^2+4)+32=0 \n\hbox{substitute a for } x^2+4: \na^2-12a+32=0 \na^2-4a-8a+32=0 \na(a-4)-8(a-4)=0 \n(a-8)(a-4)=0 \na-8=0 \ \lor \ a-4=0 \na=8 \ \lor \ a=4 \n \n\hbox{substitute 8 and 4 for a and solve for x:} \na=8 \n\Downarrow \n8=x^2+4 \ \ \ |-4 \n4=x^2 \nx=-2 \ \lor \ x=2 \n \na=4 \n\Downarrow \n4=x^2+4 \ \ \ |-4 \n0=x^2 \nx=0 \n \n\boxed{x=-2 \hbox{ or } x=0 \hbox{ or } x=2}$

The solutions for x are -2, 0, 2.

Answer:

-2,0,2

Step-by-step explanation:

The given equation is:

$(x^2+4)^(2)+32=12x^2+48$

⇒ $(x^2+4)^(2)+32=12(x^2+4)$

Substituting $(x^2+4)=a$ in the above equation, we get

⇒ $a^(2)+32=12a$

⇒ $a^2-12a+32=0$

⇒ $a^2-4a-8a+32=0$

⇒ $a(a-4)-8(a-4)=0$

⇒ $(a-8)(a-4)=0$

⇒ $a=8,4$

Now, $(x^2+4)=a$ , then substituting the value of a in this equation,

$x^(2)+4=8$ and $x^2+4=4$

⇒ $x^(2)+4=8$

⇒ $x={\pm}2$ and

⇒ $x^(2)+4=4$

⇒ $x=0$

Thus, the value of x are -2,0 and 2.

What best describes the number 5? write prime, composite, neither prime nor composite, or both.

Answers

The number 5 is a prime because prime can be divided evenly by 1, or itself. And it must be a whole number greater than 1.

Here is an example:

5 can only be divided evenly by 1 or 5, so it is a prime number. But 6 can be divided evenly by 1, 2, 3, and 6 so it is NOT a prime number (it is a composite number).

So 5 is a prime number.

Hope I Helped You!!!

prime nothing can go into that number

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Solve|3-v|<6 1.Write the inequality as two inequalities without absolute value. 2.Solve the inequality and write the solution set.

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Solve|3-v|<6

1.Write the inequality as two inequalities without absolute value.

2.Solve the inequality and write the solution set.