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Troyanec [42]
3 years ago
10

What is a counterexample of the statement “all square roots are irrational”

Mathematics
1 answer:
skad [1K]3 years ago
3 0

Rather than trying to guess and check, we can actually construct a counterexample to the statement.

So, what is an irrational number? The prefix "ir" means not, so we can say that an irrational number is something that's not a rational number, right? Since we know a rational number is a ratio between two integers, we can conclude an irrational number is a number that's not a ratio of two integers. So, an easy way to show that not all square roots are irrational would be to square a rational number then take the square root of it. Let's use three halves for our example:

\sqrt{(\frac{3}{2})^2}=\\\sqrt{\frac{9}{4}}=\\\frac{3}{2}

So clearly 9/4 is a counterexample to the statement. We can also say something stronger: All squared rational numbers are not irrational number when rooted. How would we prove this? Well, let \frac{a}{b} be a rational number. That would mean, \frac{a^2}{b^2}, would be a/b squared. Taking the square root of it yields:

\sqrt{\frac{a^2}{b^2}}}=\\ \frac{\sqrt{a^2}}{\sqrt{b^2}}=\\ \frac{a}{b}

So our stronger statement is proven, and we know that the original claim is decisively false.

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A waffle cone has a volume of 31.254 cubic centimeters and a radius of
arlik [135]

Answer:

4.78 cm

Step-by-step explanation:

The formula for the volume of a cone of radius r and height h is

V = (1/3)πr²h.  Here, V = 31.254 cm³ and r = 2.5 cm.  

We begin by solving V = (1/3)πr²h for the unknown height, h:

         31.254 cm³              3(31.254 cm³)            93.762 cm³

h = ------------------------  =  --------------------------  =  --------------------  ≈  4.78 cm

            (1/3)πr²                       π(2.5 cm)²               19.635 cm²

5 0
3 years ago
Graph the ordered pair(4,6) and its reflection over the y-axis name the coordinates for the reflection
Veseljchak [2.6K]

The new point after a reflection over the y-axis is (-4,6)

3 0
3 years ago
A factory makes 640brushes in 4 hours.Does the equation y=150x model the number of brushes made each hour?
adelina 88 [10]

Answer:

No it doesn't

Step-by-step explanation:

The equation we have is

Y = 150x

We have x = 4 hours

This is a simple question. We have to plug in the x = 4 in the equation to find out if we would have 640 brushes. If y = 640, then it models it. But if y is not equal to 640, it doesn't.

So let's solve this to see for ourselves.

Y = 150(x)

We put the value of x

Y = 150(4)

Y = 600

So we can see that the equation does not model the number of brushes made each hour. This equation models 600 brushes and not 640.

7 0
3 years ago
HELP HELP 10 POINTS AND BRAINLIEST
4vir4ik [10]
B

The square was reflected over the y axis and then was dilated 1/2
3 0
4 years ago
A square park measures 170 feet along each side. Two paved paths run from each corner to the opposite corner and extend 3 feet i
Cerrena [4.2K]

Answer:

The total area, in square feet, taken by the paths is 2,004

Step-by-step explanation:

see the attached figure with lines to better understand the problem

I can divide the figure into four right  triangles, one small square and four rectangles

step 1

Find the area of the right triangle of each corner of the path

The area of the triangle is

A=(1/2)(b)(h)

substitute the given values

A=(1/2)(3)(3)=4.5\ ft^2

step 2

Find the hypotenuse of the right triangle

Applying Pythagoras Theorem

Let

d -----> hypotenuse of the right triangle

d^{2}=3^{2}+3^{2}

d^{2}=18

d=\sqrt{18}\ ft

simplify

d=3\sqrt{2}\ ft  

The hypotenuse of the right triangle is equal to the width of the path

step 2

Find the area of the small square of the path

The area is

A=b^{2}

we have

b=3\sqrt{2}\ ft  ----> the width of the path

substitute

A=(3\sqrt{2})^{2}

A=18\ ft^2

step 3

Find the length of the diagonal of the square park

Applying Pythagoras Theorem

Let

D -----> diagonal of the square park

D^{2}=170^{2}+170^{2}

D^{2}=57,800

D=\sqrt{57,800}\ ft

simplify

D=170\sqrt{2}\ ft  

step 4

Find the height of each right triangle on each corner

The height will be equal to the width of the path divided by two, because is a 45-90-45 right triangle

h=1.5\sqrt{2}\ ft  

step 5

Find the area of each rectangle of the path

The area of rectangle is A=LW

we have

W=3\sqrt{2}\ ft ----> width of the path

Find the length of each rectangle of the path

L=(D-2h-d)/2

where

D is the diagonal of the park

h is the height of the right triangle in the corner

d is the width of the path (length side of the small square of the path)

substitute the values

L=(170\sqrt{2}-2(1.5\sqrt{2})-3\sqrt{2})/2

L=(170\sqrt{2}-3\sqrt{2}-3\sqrt{2})/2

L=(164\sqrt{2})/2

L=82\sqrt{2}\ ft

Find the area of each rectangle of the path

A=LW

we have

W=3\sqrt{2}\ ft

L=82\sqrt{2}\ ft

substitute

A=(82\sqrt{2})(3\sqrt{2})

A=492\ ft^2

step 6

Find the area of the paths

Remember

The total area of the paths is equal to the area of four right  triangles, one small square and four rectangles

so

substitute

A=4(4.5)+18+4(492)=2,004\ ft^2

therefore

The total area, in square feet, taken by the paths is 2,004

3 0
3 years ago
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