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3 years ago

Binomial Coefficients (30pts) 1. Write (a + b)^2 as a polynomial in standard form. Show work (FOIL/distributive property).

Mathematics

1 answer:

Darina [25.2K]3 years ago

4 0

Answer:

$a^2 + 2ab + b^2$

Step-by-step explanation:

Given

$(a + b)^2$

Required

Express as a standard polynomial

We have:

$(a + b)^2$

Rewrite as:

$(a + b)(a + b)$

Open bracket

$a^2 + ab + ab + b^2$

Add like terms

$a^2 + 2ab + b^2$

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Micheal solved this inequality as shown: step 1: -6(x + 3) + 10 < -2 step 2: -6x - 18 + 10 < -2 step 3: -6x - 8 < -2 st

Anastaziya [24]

<h3>Answer: addition property of inequality</h3>

================================================

Explanation:

These are the steps to focus on

step 3: -6x - 8 < -2

step 4: -6x < 6

The move from the third step to the fourth step has us adding 8 to both sides. Therefore, we use the addition property of inequality.

That property has four forms

If $a > b$ then $a + c > b+c$
If $a < b$ then $a+c < b+c$
If $a \ge b$ then $a + c \ge b+c$
If $a \le b$ then $a+c \le b+c$

It's similar to the idea of starting with a = b, then adding c to both sides to get a+c = b+c

We add the same thing to both sides to keep things balanced.

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2 years ago

I also need help with this one

solmaris [256]

The amount of pizza Mark would have is
$\frac{1}{6} + \frac{1}{4}$
Find the common denominators first. 2/12 + 3/12 = 5/12. Two slices of the 6 slice pizza is 1/3 or 4/12.

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3 years ago

If 3/7m= 7/3 n, then what is the value of m/n?

prisoha [69]

3/7m = 7/3n

9n = 49m

m/n = 9/49

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3 years ago

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Find the slope of the given points (3,4) and (7,-4)

mr Goodwill [35]

Answer:

Step-by-step explanation:

$\bold{slope\, (m)=\dfrac{change\ in\ Y}{change\ in\ X}=\dfrac{y_2-y_1}{x_2-x_1}}$

(3, 4) ⇒ x₁ = 3, y₁ = 4

(7, -4) ⇒ x₂ = 7, y₂ = -4

So the slope:

$\bold{m=\dfrac{-4-4}{7-3}=\dfrac{-8}4=-2}$

3 0

3 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E <span>to the system of inequalities from part A, and check if its is true, what we get.
</span>
$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to <span>our system of inequalities.
</span>
Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).

8 0

3 years ago

Read 2 more answers