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just olya [345]
2 years ago
10

A recipe calls for 2 tsp of oatmeal to make 7 protein bars.

Mathematics
1 answer:
harina [27]2 years ago
5 0

Answer:

3 1/3 is the answer im pretty sure

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7,000 ones is the answer
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3 years ago
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A metal conduit will be used as a pathway for wiring through a concrete block. The conduit is a 5 foot long rod with an outer di
kolbaska11 [484]

Answer:  586.59 cubic centimeters .

Step-by-step explanation:

As per given . we have

Inner diameter =  1.8 inches

⇒Inner radius :r = 0.9 in.  (radius is half of diameter)

= 0.9 x (2.54) = 2.286 cm  [∵ 1 in . = 2.54 cm]

Outer diameter = 2 inches

⇒Outer radius : R = 1 inch  = 2.54 cm

Height : h = 5 feet = 5 x(30.48) = 152.4 cm   [∵  1 foot = 30.48 cm]

The formula to find the volume of a hollow cylinder :

V=\pi(R^2-r^2)h , where R= outer radius , r= inner radius and h= height.

Now , the volume of metal in the conduit :

V=(3.14)(( 2.54)^2-(2.286)^2)(152.4)

V=(3.14)(6.4516-5.225796)(152.4)

V=(3.14)(1.225804)(152.4)

V=586.591342944\approx586.59\ cm^3

Hence, the volume of metal in the conduit is 586.59 cubic centimeters .

5 0
3 years ago
Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

Download docx
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The temperature in Scottsdale increased by 2 degrees in 2 days. If the temperature changed by the same
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