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3 years ago

Rewrite as a simplified fraction

Mathematics

2 answers:

Orlov [11]3 years ago

8 0

Answer is 7/10 (7 over 10)

Feliz [49]3 years ago

5 0

7/10 because In 0.7 we will change the decimal to fraction. First we will write the decimal without the decimal point as the numerator. Now in the denominator, write 1 followed by one zeros as there are 1 digit in the decimal part of the decimal number. Therefore, we observe that 0.7 (decimal) is converted to 7/10 fraction

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An animal feed to be mixed from soybean meal and oats must contain at least 168 lb of protein, 27 lb of fat, and 14 lb of minera

atroni [7]

Answer:

The animal farm should buy 1.775 bags of soybeans and 1.575 bags of oats

$Cost = 51.275$

Step-by-step explanation:

The given parameters can be summarized as:

$\begin{array}{cccc}{} & {x} & {y} & {Total} & {Protein} & {70} & {21} & {168} &{Fats}& {9} & {7} & {27} & {Minerals} & {7} & {1} & {14}& {Cost} & {21} & {7} & {} \ \end{array}$

Where: x = Soybeans and y = Oats.

So, the system of equations are:

$70x + 21y = 168$

$9x +7y = 27$

$7x + y = 14$

$Cost = 21x + 7y$

The best way to solve this, is using graph

Plot the following equations on a graph, and get the points of intersection:

$70x + 21y = 168$

$9x +7y = 27$

$7x + y = 14$

From the attached graph, we have:

$(x_1,y_1) = (1.636,2.545)$

$(x_2,y_2) = (1.775,1.575)$

$(x_3,y_3) = (2.023,1.256)$

Substitute each of the values of x's and y's in the cost function to get the minimum cost:

$Cost = 21x + 7y$

$(x_1,y_1) = (1.636,2.545)$

$Cost = 21 * 1.636 + 7 * 2.545$

$Cost = 52.171$

$(x_2,y_2) = (1.775,1.575)$

$Cost = 21 * 1.775 + 7 * 1.575$

$Cost = 48.3$

$(x_3,y_3) = (2.023,1.256)$

$Cost = 21 * 2.023 + 7 * 1.256$

$Cost = 51.275$

The values of x and y that gives the minimum cost is:

$(x_2,y_2) = (1.775,1.575)$

and the minimum cost is:

$Cost = 48.3$

Hence, the animal farm should buy 1.775 bags of soybeans and 1.575 bags of oats

5 0

3 years ago

Please help _____________ ______ ___________ ______

Paha777 [63]

i cant see anything???????

7 0

3 years ago

What is 1/6 more than 2/6

kramer

3/6 or 1/2

could please check out my questions? thx!

3 0

3 years ago

A giant tank in a shape of an inverted cone is filled with oil. the height of the tank is 1.5 metre and its radius is 1 metre. t

skad [1K]

The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate

of dripping of 110 cm³/s gives the following values.

1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m/s

<h3>How can the rate of change of the radius & height be found?</h3>

The given parameters are;

Height of the tank, h = 1.5 m

Radius of the tank, r = 1 m

Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s

$1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h$

From the shape of the tank, we have;

$\dfrac{h}{r} = \dfrac{1.5}{1}$

Which gives;

h = 1.5·r

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}$

$\dfrac{d}{dr} V =\dfrac{d}{dr} \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi \cdot r^2$

$\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}$

$\dfrac{dV}{dt} = 0.00011$

Which gives;

$\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi \cdot r^2}}$

When r = 0.5 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.5^2} \approx 9.34 \times 10^{-5}$

The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s

2) When the height is 20 cm, we have;

h = 1.5·r

$r = \dfrac{h}{1.5}$

$V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}$

r = 20 cm ÷ 1.5 = $13.\overline3$ cm = $0.1\overline3$ m

Which gives;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.1 \overline{3}^2} \approx \mathbf{1.313 \times 10^{-3}}$

$\dfrac{d}{dh} V = \dfrac{d}{dh} \left(\dfrac{4}{27} \cdot \pi \cdot h^3 \right) = \dfrac{4 \cdot \pi \cdot h^2}{9}$

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }$

$\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi \cdot h^2}{9}}}$

When the height is 20 cm = 0.2 m, we have;

$\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}$

The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s

3) The volume of the slick, V = π·r²·h

Where;

h = The height of the slick = 0.1 cm = 0.001 m

Therefore;

V = 0.001·π·r²

$\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}$

$\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi \cdot r}}$

When the radius is 10 cm = 0.1 m, we have;

$\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi \times 0.1} \approx \mathbf{0.175}$

The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m

Learn more about the rules of differentiation here:

brainly.com/question/20433457

brainly.com/question/13502804

3 0

3 years ago

Find the center and radius of the circle with the equation: a. center: (-5, 1) radius: 4 c. center: (-5, 1) radius: 2 b. center:

damaskus [11]

Answer:

Step-by-step explanation:

if center= (h,k)

radius=r

eq. of circle is (x-h)^2+(y-k)^2=r^2

center= (-5,1)

radius=4

eq. of circle is (x+5)^2+(y-1)^2=4^2

center=(5,-1)

radius=4

eq. of circle is (x-5)^2+(y+1)^2=4^2

center=(-5,1)

radius=2

eq. of circle is (x+5)^2+(y-1)^2=2^2

center=(5,-1)

radius=2

eq. of circle is (x-5)^2+(y+1)^2=2^2

4 0

4 years ago