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Anon25 [30]
3 years ago
13

Evaluate b^2 + 5a – 20

Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0

Answer: it cant even be factored

Step-by-step explanation:

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Evan paid $60.54 for 3 notebooks and a calculator. If the calculator costs $45.87, how much does each notebook cost? Write a two
Aleonysh [2.5K]

Answer:   4.89$

Step-by-step explanation:

60.54 is starting amount then subtract 45.87 left with 14.67 then divide by 3 get 4.89

60.54-45.87=14.67

14.67/3=4.89

7 0
3 years ago
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At the start of "Rip Van Winkle," why is Rip popular with the children of the village? A. He gives them rides in his wagon. B. H
Lyrx [107]
I believe from reading it would be C teaches them to fly kites and shoot marbles. 
4 0
3 years ago
On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.
9966 [12]

Answer:

The correct option is;

No, the distance from (0, 0) to (2, √6) is not 3 units

Step-by-step explanation:

The given parameters of the question are as follows;

Circle center (h, k) = (0, 0)

Point on the circle (x, y) = (3, 0)

We are required to verify whether point (2, √6) lie on the circle

We note that the radius of the circle is given by the equation of the circle as follows;

Distance \, formula = \sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}}

Distance² = (x - h)² + (y - k)² = r² which gives;

(3 - 0)² + (0 - 0)² = 3²

Hence r² = 3² and r = 3 units

We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;

\sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}} = Distance

Therefore;

(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²

\sqrt{\left (2-0 \right )^{2} + \left (\sqrt{6} -0 \right )^{2}} = 10

Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10

Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point  (2, √6), does not lie on the circle.

4 0
3 years ago
Read 2 more answers
What is 96 divided by 7 please help is is due by midnight
mojhsa [17]

Answer:

13.7142857143

Step-by-step explanation:

so just 13

8 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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