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2 years ago

Using only an addition, how do you add eight 8’s and get the number 1000?

Mathematics

2 answers:

Contact [7]2 years ago

5 0

Answer: 8 added 125 times=1000. 888+88+8+8+8=1000.

Step-by-step explanation:

frosja888 [35]2 years ago

4 0

Answer:

888+88+8+8+8=1000

Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

Farmer brown had ducks and horses one day he noticed that all of the Animals had a total of 12 heads and 32 feet how many animal

brilliants [131]

Answer:

4 horses and 8 ducks.

Step-by-step explanation:

Let the number of ducks be d and the number of horses be h.

d + h = 12

Now ducks have 2 feet and horses have 4 so:

2d + 4h = 32

Multiply the first equation by 2

2d + 2h = 24

Subtract this equation from the second equation:

4h - 2h = 32 - 24

2h = 8

h = 4.

So there were 4 horses and 12 - 4 = 8 ducks.

4 0

3 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 $x_2$ .