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Firdavs [7]
3 years ago
13

fiona draws a 24 centimeter rectangle.gregory draws a 24 square inch rectangle.whose rectangle is in area .how do you know

Mathematics
1 answer:
Valentin [98]3 years ago
6 0

Answer:

Therefore we come to a conclusion that,

Gregory's Rectangle is in Area.

Step-by-step explanation:

Given:

Fiona draws a 24 centimeter rectangle.

Gregory draws a 24 square inch rectangle.

To Find:

Whose rectangle is in area = ?

Solution:

Rectangle has a dimension as

Length and Width which are in meters, centimeters, inches, feet, units, etc.

For Area

Area\ of\ Rectangle=Length\times Width

here units are getting Multiplied so it becomes

unit × unit = unit²  = square units

m × m = m² =square meter

inch × inch = inch² =square inches

Hence Area is calculated in square units.

Therefore we come to a conclusion that,

Gregory's Rectangle is in Area.

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Which of the following equations is quadratic?
Dominik [7]

Answer:

C

Step-by-step explanation:

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6 0
3 years ago
Operations with rational expressions
Scorpion4ik [409]

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}

Solution:

Given expression:

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}

To solve the given expression:

First simplify: \frac{50-2 w^{2}}{3 w^{2}+9 w-30}

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30}=-\frac{2(w+5)(w-5)}{3(w-2)(w+5)}

Cancel the common factor (w + 5).

                      $=-\frac{2(w-5)}{3(w-2)}

Now substitute this in the given expression.

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{2(w-5)}{3(w-2)} \cdot \frac{w^{2}+5 w-14}{6 w-30}

Multiply the fractions \frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{3(w-2)(6 w-30)}

Factor the denominator 3(w-2)(6 w-30) =18(w-2)(w-5)

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{18(w-2)(w-5)}                        

Cancel the common factor 2(w – 5).

                               $=-\frac{w^{2}+5 w-14}{9(w-2)}

Factor the numerator w^{2}+5 w-14=(w-2)(w+7)

                               $=-\frac{(w-2)(w+7)}{9(w-2)}

Cancel the common factor (w – 2).

                               $=-\frac{w+7}{9}

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}

3 0
3 years ago
Base on the data what is the probability of rolling a 3 or a 5
Kobotan [32]

Answer:

1/3

Step-by-step explanation:

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3 0
4 years ago
Read 2 more answers
Arrange the equations in the correct sequence to rewrite the formula for displacement, , to find a. In the formula, d is displac
OverLord2011 [107]

Answer:

2(d-vt)=-at^2

a=2(d-vt)/t^2

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Step-by-step explanation:

Arrange the equations in the correct sequence to rewrite the formula for displacement, d = vt—1/2at^2 to find a. In the formula, d is

displacement, v is final velocity, a is acceleration, and t is time.

Given the formula for calculating the displacement of a body as shown below;

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Where,

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v = final velocity

a = acceleration

t = time

To make acceleration(a), the subject of the formula

Subtract vt from both sides of the equation

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d - vt=vt - vt - 1/2at^2

d - vt= -1/2at^2

2(d - vt) = -at^2

Divide both sides by t^2

2(d - vt) / t^2 = -at^2 / t^2

2(d - vt) / t^2 = -a

a= -2(d - vt) / t^2

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3 years ago
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Deffense [45]

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By doing this, you get :

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So you get

3

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8 0
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