State the order of operations in -2(x-5)^2+8: 1. 2. 3.

2 answers:

4 0

You would use PEMDAS

P stands for parentheses, x-5 can’t be simplified, so on to E.

E stands for exponent so next you would square x-5 getting x^2-10x+25.

M stands for multiplication, you multiply your result from E by -2. So, -2(x^2-10x+25)+8. And then -2x^2+20x-50+8.

D stands for division, and there is none so moving on.

A stands for addition, and so we add next. -2x^2+20x-42.

You can’t simplify any further so there you go!
Hope this helps!

zlopas [31]3 years ago

4 0

Hey there!

Just distribute:
2(x - 5)^2 + 8
= 2x^2 + 20x - 50 + 8
Combine your like terms
-50 & 8
= -50 + 8 ➡️ -42

Answer: 2x^2 + 20x + -42

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The temperature in fargo was -28 degrees on monday afternoon. by the evening it dropped an additional 15 degrees. what was the t

sashaice [31]

Step by step explanation: -28-15

When adding negative with positive it will always be negative so for this question, subtract -28 with 15 which would be -43. So -43 would be the temperature in the evening.

7 0

2 years ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .