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2 years ago

Ricardo is building a new garden and is buying soil to fill it up. he buys 4 kilograms of soil for $10.00.

Mathematics

2 answers:

likoan [24]2 years ago

5 0

Answer:

math duh

Step-by-step explanation:

ExtremeBDS [4]2 years ago

4 0

10/4=$2.5. Is the answer

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The Japanese Yen corresponds to the American Dollar. If the exchange rate is 150 Yen for 1 Dollar, then a Japanese car sold in A

Papessa [141]

Answer:

125,000

Step-by-step explanation:

Take the Japanese yen and divide by 12

1,500,000 divided by 12 = 125,000

3 0

3 years ago

Whats the correct answer <br> 41/8928

chubhunter [2.5K]

The answer to 41 divided by 8,928 would be 0.0046

6 0

2 years ago

Daniel [21]

Answer:

Step-by-step explanation:

P = (-4, 3)

Reflecting across the y axis negates the x value

P₁ = (4, 3)

Reflecting across the x axis negates the y value

P' = (4, -3)

option D

4 0

2 years ago

Determine if 9x^2-42x+49 can be the area of a square. If so, what would the value of X be if the square is 64 meters? A=s^2

Anton [14]

I don't know if it could be the area of a square but x has to be 8.34. (ROUNDED)

8 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 $x_2$ .

Remembering,

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

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

2 years ago