Problem

$3x-12>33^{+}$ 1 point
$○\left(x>-15\right)$
$○\left(x>15\right)$
$○\left(x<-15\right)$
$○\left(x<15\right)$
Solve: $|p+2|+4<10≠$ 1 point
$-8<p<4$
$8<p<-4$
$-8<p<-4$
$8<p<4$
Which of the following is equivalent $to|x-12|=147≠$ 1 point
$\left(2,26\right)$
$\left(2,-26\right)$
$\left(-2,26\right)$
$\left(-2,-26\right)$
$3a-2=10≠$ 1 point
$a=2$
$a=4$
a = 6
O
a = 8
O O This is a required question

7th-9th grade

Algebra

Question content

help me plsSolution

Qanda teacher - Shashi

You can also ask preffered match Questions to me.

Student

Thanks

Still don't get it?

Ask this question to Qanda teacherSimilar problem

$13.$ Which of the following is a factor
of $x^{A}4+x^{A}2-x^{n}2-x-187$ $x$
$○$ a. $x-2$
$○$ b. $x-5$
$○$ O c. $x+7$
$○$ d. $x+5$
$Other^{.}$
This is a required question

10th-13th grade

Algebra