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1 year ago

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A french fry stand at the fair serves their fries in paper cones. The cones have a radius of 222 inches and a height of 666 inch

es. It is a challenge to fill the narrow cones with their long fries. They want to use new cones that have the same volume as their existing cones but a larger radius of 444 inches.
What will the height of the new cones be?

1 answer:

rewona [7]1 year ago

4 0

Based on the given parameters about old cone, the height of the new cone is 12 inches

<h3>Equivalent ratio</h3>

Old cone:

Radius, r = 2 inches
Height, h = 6 inches

New cone:

Radius, r = 4 inches
Height, h = h

equate ratio of radius to height in old and new cone

2 : 6 = 4 : h

2/6 = 4/h

cross product

2 × h = 6 × 4

2h = 24

h = 24/2

h = 12 inches

Learn more about ratio:

brainly.com/question/2328454

#SPJ1

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tia_tia [17]

Answer:

$8-6i$

Step-by-step explanation:

Multiply out $(3-i)^2$ in order to find out which expression it is equivalent to.

1) Since the whole quantity is squared, write it out as $(3-i)(3-i)$ .

$(3-i)^2\\(3-i)(3-i)$

2) Multiply binomials by using the FOIL method. Multiply the terms that are listed first in each binomial, then the ones that are listed outermost when looking at both binomials, then innermost, and finally the last terms listed in each binomial. Simplify and combine like terms.

$(3-i)(3-i)\\9 - 3i - 3i + i^2$

$9 - 6i + i^2$

3) $i = \sqrt{-1}$ , so $i ^2$ must be $(\sqrt{-1} )(\sqrt{-1})$ . Thus, $i^2$ is equivalent to -1. Knowing this, simplify and combine like terms.

$9 - 6i + ((-\sqrt{1} )(\sqrt{-1} ))$

$9 - 6i -1\\8 - 6i$

Thus, it is equivalent to $8 - 6i$ .

6 0

3 years ago

Please Help it urgent

denis-greek [22]

Answer:

Angle BEA

Step-by-step explanation:

They share the same straight line and their degrees add to 180

7 0

3 years ago

The endpoints of a line segment graphed on a Cartesian coordinate system are (4, 1) and (-2, -4). What are the coordinates of th

Masja [62]

The midpoint of (x1,y1) and (x2,y2) is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

just average them

(4,1) and (-2,4)
the midpoint is $(\frac{4-2}{2},\frac{1-4}{2})=(\frac{2}{2},\frac{-3}{2})=$ (1,-1.5)

5 0

3 years ago

Read 2 more answers

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:

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

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2 years ago

Describe the graph of the function.y = |x – 4| – 7

juin [17]

This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:

(x-4)=0 ===> x=4,

so that now you have to plot 2 functions!

For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:

|x-4| = -(x-4)=4-x
Then:

for x<=4, y = -x+4-7 = -x-3

for x=>4, (x-4) is positive, so no changes:

y= x-4-7 = x-11,

Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11

In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:

x=4 and x= 3 for y=-x-3

And just x=5 for y=x-11

The reason is that the absolute value is continuous, so x=4 works for both:

x=4===> y=-4-3 = -7

x==4 ====> y = 4-11=-7!

abs() usually have a cusp int he point where it is =0

Hope it helps, despite being this long!

3 0

2 years ago

Read 2 more answers