| 3x - 1 | + 7 = 2
| 3x - 1 | = 2 - 7
| 3x - 1 | = -5.....no solution because an absolute value cannot equal a negative number.

| 2x + 1 | + 4 = 3
| 2x + 1 | = 3 - 4
|2x + 1 | = -1...same thing...no solution..cannot equal a neg. number

so Eq. 1 and Eq. 2 both have NO SOLUTIONS :)

8 0

4 years ago

Read 2 more answers

Will give brainliest no files will report.

Phantasy [73]

Answer: i don't know you have to figure it out yourself duh.

Step-by-step explanation: this is none of your beeswax ok

4 0

3 years ago

What is the area of this figure?<br><br> Enter your answer in the box.

Colt1911 [192]

IF they want to know the area you multiply all of them

16 x 20 x 8

You will get 2560

4 0

4 years ago

when the polynomial f(x) is divided by (x-2) the remainder is 4, and when it is divided by (x-3) the remainder is 7. Given that

natima [27]

$f(x)=(x-2)(x-3)Q(x)+ax+b$

Recall the polynomial remainder theorem: the remainder upon dividing a polynomial $p(x)$ by $x-c$ is equal to $p(c)$ . This means that $f(2)=4$ and $f(3)=7$ , which tell us

$4=2a+b$

$7=3a+b$

From here we can solve for $a,b$ :

$4=2a+b\implies b=4-2a$

$7=3a+b=3a+(4-2a)\implies a=3\implies b=-2$

so that

$f(x)=(x-2)(x-3)Q(x)+3x-2$

Now,

$\dfrac{f(x)}{(x-2)(x-3)}=Q(x)+\dfrac{3x-2}{(x-2)(x-3)}$

so the remainder upon dividing $f(x)$ by $(x-2)(x-3)$ is $3x-2$ .

Next, if $f$ is a cubic function, then $Q(x)$ is a linear polynomial that can be written as $Q(x)=cx-d$ . The coefficient of $x^3$ in $f(x)$ is 1 (unity), so that expanding $f(x)$ gives us