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Flauer [41]
2 years ago
6

Evalue f(x)=(-2)^2-3(-2)+5/(-2)^2+1

Mathematics
1 answer:
Nina [5.8K]2 years ago
8 0
F(x)=4+6+5/4+1
F(x)=4+6+5/5
F(x)=10+1=11

Answer =11
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37/50 of a number is what percentage of that number
Ronch [10]

Answer:

74%

Step-by-step explanation:

Multiply both the numerator and denominator by 2 so that you will have a number over 100. After doing this you will have 74/100 which is 74%

6 0
3 years ago
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Write in slope-intercept form an equation of a line that passes through the point (8, 2) that is perpendicular to the graph of t
Citrus2011 [14]

Answer:

y= -1/4 x + 4

5 0
3 years ago
When Matthew eats a pizza he always eats a whole slice at a time with no bites left over. Which graph best represents this situa
RoseWind [281]

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b

Step-by-step explanation:

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In sunlight, a vertical stick has a height of 4 ft and casts a shadow 2 ft long at the same time that a nearby tree casts a shad
marshall27 [118]
Remark
Make up a proportion that relates the tree and its properties to the stick and its properties.

Givens
Stick height (s_h) = 4
Stick shadow(s_s) = 2
Tree shadow = t_s = 18
Tree height = x

Formula
s_h / s_s = x / t_s

Substitute and solve.
4/2 = x / 18  Cross multiply
4*18 = 2 * x
72 = 2x        Divide by 2
72/2 = x       
x = 36

The height of the tree is 36 feet.     
6 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:

  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})

5 0
2 years ago
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