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4 years ago

A ratio compares two fractions. True or false

Mathematics

2 answers:

Ostrovityanka [42]4 years ago

5 0

It is true because I searched it up

Oxana [17]4 years ago

3 0

It is true. A ratio compares to two fractions

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Answers for question 3 and 4??

IrinaVladis [17]

Answer:

i need this toooooo

Step-by-step explanation:

7 0

4 years ago

What is the image of A(8,2) under R90 ?<br> A. (-2,8)<br> B. (2,8)<br> C.( 8,-2)<br> D.(-8,2)

swat32

Answer: A. (-2,8)

Step-by-step explanation:

We know that the rule for rotation of $R_{90}$ is given by :-

$(x,y)\rightarrow(-y,x)$ , where (x,y) is the point of pre-image and (-y,x) is a point of image.

The given point of pre-image : A(8,2)

Then after rotation of $R_{90}$ , we have the image point as :-

$(8,2)\rightarrow(-2,8)$

Hence, the image point of A(8,2) is (-2,8) .

5 0

3 years ago

Awnser please fast ,,,,,,,,,

leva [86]

Answer:

there's this one app called Photomatix that does any algebra problem

4 0

3 years ago

A piece of wire of length 5050 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the

algol [13]

Answer:

x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

Step-by-step explanation:

Let x be the length of wire that is cut to form a circle within the 5050 wire, so 5050 - x would be the length to form a square.

A circle with perimeter of x would have a radius of x/(2π), and its area would be

$A_c = \pir^2 = \pi (\frac{x}{2pi})^2 = \frac{x^2}{4\pi}$

A square with perimeter of 5050 - x would have side length of (5050 - x)/4, and its area would be

$A_s = (\frac{5050 - x}{4})^2 = \frac{(5050 - x)^2}{16}$

The total combined area of the square and circles is

$\sum A = A_c + A_s = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$

To find the maximum and minimum of this, we just take the 1st derivative, and set it to 0

$A' = \frac{2x}{4\pi} + \frac{-2(5050-x)}{16} = 0$

$\frac{x}{2\pi} - \frac{5050 - x}{8} = 0$

Multiple both sides by 8π and we have

$4x - 5050\pi + x\pi = 0$

$x(4 + \pi) = 5050\pi$

$x = \frac{5050\pi}{4 + \pi} = 2221.5$

At x = 2221.5:

$A = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$ = 392720 + 500026 = 892746 [/tex]

At x = 0, $A = 5050^2/16 = 1593906$

At x = 5050, $A = \frac{5050^2}{4\pi} = 2029424$

As 892746 < 1593906 < 2029424, x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

3 0

3 years ago

What is 65.85 rounded to the nearest tenth

Alinara [238K]

65.9 yo. 5 and above give it a shove 4 and beliw let it go

5 0

3 years ago

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