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pickupchik [31]
2 years ago
12

Can someone help? I suck at this

Mathematics
1 answer:
Usimov [2.4K]2 years ago
4 0

Answer:

23 cm squared

Step-by-step explanation:

Split the shape into two shapes.

<u>Shape one:</u>

5 x 3 = 15 cm squared

<u>Shape two:</u>

7 - 3 = 4 (length of a side of the second shape)

4 x 2 = 8 cm squared

15 + 8 = 23 cm squared

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Find the area of shape 1 and shape 2
kirza4 [7]
You can use Pick's theorem, where:

Area = Number of interior points (points inside the polygon) + \dfrac{1}{2} · Number of boundary points (points on the polygon's perimeter) - 1

so:

1.Yellow:

Interior points = 3
Boundary points = 8

A=3+\dfrac{1}{2}\cdot8-1=3+4-1=\boxed{6\text{ units}}

2. Red:

Interior points = 2
Boundary points = 10

A=2+\dfrac{1}{2}\cdot10-1=2+5-1=\boxed{6\text{ units}}
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3 years ago
Given that y varies directly as x², find the percentage increase in y when x doubles.
stepladder [879]

Answer:400%

Step-by-step explanation: by the commutativity of multiplication we find (2x)^2=4(x^2)=4(y)

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1 year ago
The graph Y= F(x), what is f (-4)
guapka [62]

Answer:2

Step-by-step explanation:

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3 years ago
What is the direct variation equation for the table of ordered pairs
ivolga24 [154]
Y = kx
4.8 = 4k
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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}

                           

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6 0
3 years ago
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