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Ganezh [65]
3 years ago
5

Find the eighth term of an arithmetic sequence whose fourth term is 19 and whose fifteenth is 52.

Mathematics
2 answers:
Ganezh [65]3 years ago
7 0

Answer:

The correct answer is actually B, 31

Step-by-step explanation:

Here is a link to a video giving an in depth explanation, just delete all of the (space)s

https://( )www.numerade( ).com/questions/find-the-( )indicated-term-( )of-each-sequence-the( )-eighth-term-of-( )the-arithmetic-( )sequence-whose-( )fourth-ter/

Elina [12.6K]3 years ago
3 0

Answer:

its C can u give me brainly

Step-by-step explanation:

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consider the graph of the following qudratic equation. y= -x^2 -10x + 24.What is the y-value of the vertex?
3241004551 [841]

Given that the quadratic equation is y=-x^{2}-10 x+24

We need to determine the y - value of the vertex.

<u>The x - value of the vertex:</u>

The x - value of the vertex can be determined using the formula,

x=-\frac{b}{2 a}

where a=-1, b=-10, c=24

Substituting these values, we get;

x=-\frac{(-10)}{2(-1)}

Simplifying the terms, we get;

x=-\frac{-10}{-2}

x=-5

Thus, the x - value of the vertex is -5.

<u>The y - value of the vertex:</u>

The y - value of the vertex can be determined by substituting the x - value of the vertex ( x = -5) in the equation y=-x^{2}-10 x+24

Thus, we get;

y=-(-5)^{2}-10(-5)+24

Simplifying the values, we have;

y=-25+50+24

y=49

Thus, the y - value of the vertex is 49.

8 0
3 years ago
Cylinder A has a radius of 10 inches and a height of 5 inches. Cylinder B has a volume of 750π. What is the percentage change in
Firlakuza [10]

Answer:

50% change in volume

Step-by-step explanation:

<h2>This problem bothers on the mensuration of solid shapes.</h2>

In this problem we are to find the volume of the first  cylinder and compare with the second cylinder.

Given data

Volume v =  ?

Diameter d= ?

Radius r =  10 in

Height h=  5 in

we know that the volume of a cylinder is expressed as

volume = \pi r^{2}h

Substituting our given data we have

volume = \pi*10^{2}*5\\ volume= \pi *100*5\\volume= 500\pi in^{3} \\

The first cylinder as a volume of 500\pi

The change in volume is 750\pi - 500\pi = 250\pi

percentage = \frac{250\pi }{500\pi } *100

percentage=0.5*100= 50%

6 0
3 years ago
Read 2 more answers
A line passes through the point (-4, -2) and has a slope of -5/2. Write an equation in slope- intercept form for this line.
kirill115 [55]

Answer:

<h2>           y = -⁵/₂x - 12 </h2>

Step-by-step explanation:

The point-slope form of the equation is y - y₀ = m(x - x₀), where (x₀, y₀) is any point the line passes through and m is the slope:

m = -⁵/₂

(-4, -2)    ⇒   x₀ = -4,  y₀ = -2

The point-slope form of the equation:

y + 2 = -⁵/₂(x + 4)

So:

y + 2 = -⁵/₂x - 10         {subtract 2 from both sides}

y = -⁵/₂x - 12             ←  the slope-intercept form of the equation

5 0
3 years ago
Sole for |p+8| over 2 =5
Lubov Fominskaja [6]

Answer:

p =2                     p=-18

Step-by-step explanation:

|p+8|

-------------- = 5

2

Multiply each side by 2

|p+8|

-------------*2 = 5*2

2

|p+8|  =10

There is a positive and a negative solution

p+8=10         p+8=-10

Subtract 8 from each side

p+8-8=10-8         p+8-8=-10-8

p =2                     p=-18



7 0
4 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
6 0
3 years ago
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