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3 years ago

100 is smaller than 110 by how many percent?

2 answers:

8 0

Compare 100 to 110, and see that to go from 100 to 110 is a 10% increase, but to go from 110 back down to 100 is a 9.09% decrease not a 10% decrease

Bad White [126]3 years ago

8 0

Answer:

9.09%

Any percent increase or decrease can be determined by using the process:

change/original ×100%

The change is the amount of increase or decrease, simply subtract:

Bigger value - smaller value

Step-by-step explanation:

in this case 110-100

----------- ×100

110

=100/11 = 9.09 %

hope it helps

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What would be the equation -2x+y+3=0 in y=mx+b form?

ololo11 [35]

Answer:

The answer is y = 2x - 3.

Step-by-step explanation:

Move 2x to the right side of the equation by adding it to both sides:

-2x + y + 3 = 0

+ 2x + 2x

y + 3 = 2x

Then, move 3 to the right side of the equation by subtracting it form both sides:

y + 3 = 2x

- 3 - 3

y = 2x - 3

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$1068.66

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3 years ago

Write a coordinate proof given quadrilateral ABCD with vertices A (3, 2), B (8, 2), C (5, 0), and D (0, 0).

EleoNora [17]

Answer:

Step-by-step explanation:

There is really no way to show that of I know of without drawing it but here is this. I hope this helps and not really exact on position but that is very close.

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3 years ago

Find two vectors in R2 with Euclidian Norm 1 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 $x_2$ .