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svetoff [14.1K]
3 years ago
6

How do you solve 2 2/5 divided by 1/12

Mathematics
2 answers:
AfilCa [17]3 years ago
6 0
You have to turn 2 2/5 into a improper fraction. Then divide it afterwards hold,change and flip.
lisov135 [29]3 years ago
4 0
Change 2 2/5 to an improper fraction (12/5) and multiply by the reciprocal of 1/12 (12/1) this is 144/5 then simplify to 28 4/5
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TWO LIKE FRACTIONS OF 14/18
ss7ja [257]

Answer:

7/9, 28/36

Step-by-step explanation:

14/18

Divide the numerator and denominator of 14/18 by 2: 7/9

Multiply the numerator and denominator of 14/18 by 2: 28/36

7 0
4 years ago
Read 2 more answers
Help ASAP please!! Find the length of the missing side. If necessary, round to the nearest tenth.
Aleksandr-060686 [28]

Answer:

=20.5

Step-by-step explanation:

1. use the Pythagorean theorem: a^2+ b^2=c^2  (arms+legs squared=hypotenuse)

2. plug in the numbers to the equation:

14^2+15^2=c

225+196=c^2

421=C^2

the square root of 421 is 20.5 which is the hypotenuse of the triangle

7 0
3 years ago
Does anyone know how to do this? I’m confused
nikklg [1K]

Answer:

cos(θ)

Step-by-step explanation:

Para una función f(x), la derivada es el límite de  

h

f(x+h)−f(x)

​

, ya que h va a 0, si ese límite existe.

dθ

d

​

(sin(θ))=(  

h→0

lim

​

 

h

sin(θ+h)−sin(θ)

​

)

Usa la fórmula de suma para el seno.

h→0

lim

​

 

h

sin(h+θ)−sin(θ)

​

 

Simplifica sin(θ).

h→0

lim

​

 

h

sin(θ)(cos(h)−1)+cos(θ)sin(h)

​

 

Reescribe el límite.

(  

h→0

lim

​

sin(θ))(  

h→0

lim

​

 

h

cos(h)−1

​

)+(  

h→0

lim

​

cos(θ))(  

h→0

lim

​

 

h

sin(h)

​

)

Usa el hecho de que θ es una constante al calcular límites, ya que h va a 0.

sin(θ)(  

h→0

lim

​

 

h

cos(h)−1

​

)+cos(θ)(  

h→0

lim

​

 

h

sin(h)

​

)

El límite lim  

θ→0

​

 

θ

sin(θ)

​

 es 1.

sin(θ)(  

h→0

lim

​

 

h

cos(h)−1

​

)+cos(θ)

Para calcular el límite lim  

h→0

​

 

h

cos(h)−1

​

, primero multiplique el numerador y denominador por cos(h)+1.

(  

h→0

lim

​

 

h

cos(h)−1

​

)=(  

h→0

lim

​

 

h(cos(h)+1)

(cos(h)−1)(cos(h)+1)

​

)

Multiplica cos(h)+1 por cos(h)−1.

h→0

lim

​

 

h(cos(h)+1)

(cos(h))  

2

−1

​

 

Usa la identidad pitagórica.

h→0

lim

​

−  

h(cos(h)+1)

(sin(h))  

2

 

​

 

Reescribe el límite.

(  

h→0

lim

​

−  

h

sin(h)

​

)(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)

El límite lim  

θ→0

​

 

θ

sin(θ)

​

 es 1.

−(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)

Usa el hecho de que  

cos(h)+1

sin(h)

​

 es un valor continuo en 0.

(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)=0

Sustituye el valor 0 en la expresión sin(θ)(lim  

h→0

​

 

h

cos(h)−1

​

)+cos(θ).

cos(θ)

5 0
3 years ago
Read 2 more answers
Write the correct description beneath each of
Ksju [112]

1) Inequality

2) Equation

Step-by-step explanation:

We need to identify the correct description

1) 7x+9<25

The sign < is called greater than and is an inequality symbol

So, the given term is an Inequality

2) 2x-3=5x+12

This is an equation because we equal (=) sign in it.

Solving the equation we can find value of x.

So, the given term is an Equation

Keywords: Solving Equations

Learn more about Solving Equations at:

  • brainly.com/question/1563227
  • brainly.com/question/2403985
  • brainly.com/question/11229113

#learnwithBrainly

7 0
3 years ago
The length of a rectangle is three more than twice it’s width. The perimeter of the rectangle is 246 cm. Write an equation to fi
myrzilka [38]

Answer:

83 cm

Step-by-step explanation:

Let the length of the rectangle be L cm and the width be W cm.

L= 2W +3 -----(1)

Perimeter of rectangle= 2(length) +2(width)

2L +2W= 246 -----(2)

Substitute (1) into (2):

2(2W +3) +2W= 246

Expand:

4W +6 +2W= 246

Simplify:

6W +6= 246

Divide both sides by 6:

W +1= 41

W= 41 -1

W= 40

∴ width of the rectangle is 40cm.

Substitute value of W into (1):

L= 2(40).+3

L= 80 +3

L= 83

Thus, the length of the rectangle is 83 cm.

8 0
3 years ago
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