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IRINA_888 [86]
3 years ago
5

Three students, Angie, Bradley, and Carnell, are being selected for three student council offices: president, vice president, an

d treasurer. In each arrangement below, the first initial of each person’s name represents that person’s position, with president listed first, vice president second, and treasurer third. Which shows the possible outcomes for the event? ABC ABC, BAC, CBA AAA, BBB, CCC ABC, ACB, BCA, BAC, CAB, CBA
Mathematics
2 answers:
elena55 [62]3 years ago
8 0
Does this have to be only one of them or multiple?
Morgarella [4.7K]3 years ago
4 0

Answer:

the answer is D trust me

Step-by-step explanation:

You might be interested in
Do u agree or disagree and why?
pshichka [43]

Answer:

Erm. i would disagree

Step-by-step explanation:

Because they don't look the same or anything like that.

4 0
3 years ago
Read 2 more answers
(-2,+<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D" id="TexFormula1" title="\sqrt{x}" alt="\sqrt{x}" align="absmiddle" class=
rewona [7]

(-2 = √x - 7)x² (calculate)

(-9 = √x)x² (remove the parantheses)

-9x² + √x x²

6 0
3 years ago
Identify three solutions of the graph shown. Please help
barxatty [35]
From the graph, we can see that the graph crosses the x-axis at the point (1.5, 0) the graph also passes through point (1, 1) and the graph crosses the y-axis at the point (0, 3).

Therefore, points (1.5, 0), (1, 1) and (0, 3) are some of the solutions of the graph.
5 0
3 years ago
Let $r$ and $s$ be the roots of $3x^2 + 4x + 12 = 0.$ Find $r^2 + s^2.$ Pls help.
nirvana33 [79]

Answer:

-56/9

Step-by-step explanation:

By Vieta's formulas,

$r + s = -\frac{4}{3}$ and $rs = \frac{12}{3} = 4.$ Squaring the equation $r + s = -\frac{4}{3},$ we get

$r^2 + 2rs + s^2 = \frac{16}{9}.$ Therefore,

$r^2 + s^2 = \frac{16}{9} - 2rs = \frac{16}{9} - 2 \cdot 4 = -\frac{56}{9}}$

5 0
3 years ago
The solution set for <br> 6a^2-a-5=0
Alex_Xolod [135]

Answer:

a=1 or a=5/6

Step-by-step explanation:

I'm going to attempt to factor 6a^2-a-5

a=6

b=-1

c=-5

Find two numbers that multiply to be a*c and add to be b.

a*c=-30                                   =-6(5)

b=-1                                          =-6+5

So replace -a with -6a+5a in the expression we started with

6a^2-6a+5a-5

now we factor by grouping

6a(a-1)+5(a-1)

(a-1)(6a-5)

Now let's solve the equation:

(a-1)(6a-5)=0

So a=1 or a=5/6

4 0
3 years ago
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