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Naya [18.7K]
3 years ago
15

Escreva os numeros racionais na forma de fração: +6,8,-0,375,-0,02 e 0,07

Mathematics
2 answers:
Mars2501 [29]3 years ago
5 0
Can you translate into English please :) 
sergij07 [2.7K]3 years ago
5 0
From what I can understand, it says to write the rational numbers in the form of fractions.
+6=6/10
8=8/10
or +6,8=  6 4/5
-0,375=-375/1000
-0,02=-2/100
0,07=7/100

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3 = x/12 - 4 what is x
alexgriva [62]

Answer:

x=84

Step-by-step explanation:

Add 4 on both sides

3=x/12-4

7=x/12

Multiply both sides by 12

(12)7=x/12(12)

84=x

Hope this helps! :)

3 0
3 years ago
Read 2 more answers
Solve for x: negative 3 over 2, multiplied by x minus 9 equals negative 27 −24 −12 11 12 Btw please show the work bc i not only
Firdavs [7]

Answer:

\boxed{ x = 12}

Step-by-step explanation:

<u>Given Equation is:</u>

\frac{-3}{2} (x) - 9 = -27

Adding 9 to both sides

=> \frac{-3x}{2} = -27+9

=> \frac{-3x}{2} = -18

Multiplying 2 to both sides

=> -3x = -18 * 2

=> -3x = -36

Dividing both sides by -3

=> x = 12

6 0
3 years ago
Read 2 more answers
Could you please help me on this question out really appreciate it!
PtichkaEL [24]
The answer is 24

hope this helps :)
4 0
3 years ago
The weight of an object on mars varies directly is proportional as the weight of the object on earth a 75 pound object on earth
dedylja [7]

Answer:

It weighs 40 pounds on mars

Step-by-step explanation:

Direct proportion is the relationship between two variables when their ratio is equal to a constant value

  • It can be expressed as y = k x, where y varies directly as x and k is constant of variation
  • It can be expressed as  \frac{y_{1}}{x_{1}}=\frac{y_{2}}{x_{2}}=k

∵ The weight of an object on mars varies directly as the weight

    of the object on earth

∴ w(m) = k × w(e)

∵ The weight of the object on earth is 75 pound

∴ w(e) = 75

∵ The weight of this object on mars is 25 pounds

∴ w(m) = 25

- Substitute them in the equation of variation

∵ 25 = k × 75

- Divide both sides by 75

∴ \frac{1}{3} = k

∴ w(m) = \frac{1}{3} × w(e)

∵ An object weighs 120 pounds on earth

∴ w(e) = 120

- Substitute it in the equation of variation

∴ w(m) = \frac{1}{3} × 120

∴ w(m) = 40

∴ It weighs 40 pounds on mars

5 0
3 years ago
Water is being pumped into a conical tank that is 8 feet tall and has a diameter of 10 feet. If the water is being pumped in at
Deffense [45]

The rate of change of the depth of water in the tank when the tank is half

filled can be found using chain rule of differentiation.

When the tank is half filled, the depth of the water is changing at  <u>1.213 × </u>

<u>10⁻² ft.³/hour</u>.

Reasons:

The given parameter are;

Height of the conical tank, h = 8 feet

Diameter of the conical tank, d = 10 feet

Rate at which water is being pumped into the tank, = 3/5 ft.³/hr.

Required:

The rate at which the depth of the water in the tank is changing when the

tank is half full.

Solution:

The radius of the tank, r = d ÷ 2

∴ r = 10 ft. ÷ 2 = 5 ft.

Using similar triangles, we have;

\dfrac{r}{h} = \dfrac{5}{8}

The volume of the tank is therefore;

V = \mathbf{\dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h}

r = \dfrac{5}{8} \times h

Therefore;

V = \dfrac{1}{3} \cdot \pi \cdot \left(  \dfrac{5}{8} \times h\right)^2 \cdot h = \dfrac{25 \cdot h^3 \cdot \pi}{192}

By chain rule of differentiation, we have;

\dfrac{dV}{dt} = \mathbf{\dfrac{dV}{dh} \cdot \dfrac{dh}{dt}}

\dfrac{dV}{dh}=\dfrac{d}{h} \left(  \dfrac{25 \cdot h^3 \cdot \pi}{192} \right) = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64}}

\dfrac{dV}{dt} = \dfrac{3}{5}  \ ft.^3/hour

Which gives;

\dfrac{3}{5} =  \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64} \times \dfrac{dh}{dt}}

When the tank is half filled, we have;

V_{1/2} = \dfrac{1}{2} \times  \dfrac{1}{3} \times \pi \times 5^2 \times 8 =\mathbf{ \dfrac{25 \cdot h^3 \cdot \pi}{ 192}}

Solving gives;

h³ = 256

h = ∛256

\dfrac{3}{5} \times \dfrac{64}{25 \cdot h^2 \cdot \pi} = \dfrac{dh}{dt}

Which gives;

\dfrac{dh}{dt} = \dfrac{3}{5} \times \dfrac{64}{25 \cdot (\sqrt[3]{256}) ^2 \cdot \pi} \approx \mathbf{1.213\times 10^{-2}}

When the tank is half filled, the depth of the water is changing at  <u>1.213 × 10⁻² ft.³/hour</u>.

Learn more here:

brainly.com/question/9168560

6 0
2 years ago
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