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defon
3 years ago
8

-10 I will give brainliest answer instantly if correct

Mathematics
1 answer:
Alex3 years ago
7 0

Answer:

wait what? Lol thanks for the points luv <3

Step-by-step explanation:

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A cylinder’s radius is reduced to 2004-06-01-04-00_files/i0480001.jpg its original size and the height is quadrupled. How has th
asambeis [7]
I can't answer this question if we don't know by what scale the cylinder's radius was reduced. Luckily, I found the same problem that says the radius was reduced to 2/5. So, we find the ratio of both volumes.

V₁ = πr₁²h₁
V₂ = πr₂²h₂

where r₂ = 2/5*r₁ and h₂ = 4h₁

V₂/V₁ = π(2/5*r₁ )²(4h₁)/πr₁²h₁= 8/5 or 1.6

<em>Thus, the volume has increased more by 60%.</em>
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The graph shows the number of paintballs a machine launches, y, in x seconds:
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The answer is B, 20 balls in 4 seconds. Or 5 balls per second.
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Which of these is a simplified form of the equation 7x + 4 = 5 + 3x + 1x?
svetoff [14.1K]
Your answer is going to be <span>3x=1.</span>
3 0
3 years ago
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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:

  1. v_1 = (x_1,y_1)
  2. 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})

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You will need 6/4 or 1 1/2 cups of blueberries to make 48 muffins (2 dozen),
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