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3 years ago

Which number is the common difference of the sequence -92,-74,-54,-38,-20?

2 answers:

8 0

The common difference if there is one is the constant difference that occurs between any term and the term before it.... in this case:

There is no common difference,

dx=18,20,16,18 the difference or velocity is not constant...

d2x=2, -4,2 the acceleration is not constant...

d3x=-6,6 the thrust is not constant

Now we might be tempted to say that:

d4x=12 and say that that is constant and we COULD make a quartic equation fit all the data points, but without further data points in the sequence there is no mathematical proof that the quartic equation would produce accurate data points outside of the range given...

And solving a system of five equations for five unknowns is tedious for such a problem...a^4+bx^3+cx^2+dx+e=y

madam [21]3 years ago

8 0

Answer:

The answer is 18

Step-by-step explanation:

Each number is getting 18 added to it

You might be interested in

Jules kicks a soccer ball off the ground and into the air with an initial velocity of 25 feet per second. Assume the starting he

kondor19780726 [428]

Use the formula h(t) = -16t^2 + vt + s,

where t = time in air,

v = starting upward velocity,

s = the objects starting height.

from ground (s), the starting height is 0,

the ball's initial velocity (v) 25 feet per second

Solve:

v - 16t^2 + (25)t + 0

= -16t^2 + 25t. 0

Factor:

= -16t^2 + 25t

Find GCF:

0 = -16t^2 + 25t

Divide both sides by (t)

t(-16 + 25) = 0

t = 0; -16 + 25 = 0

Now Subtract 25 from both sides of equation:

-16 = -25

divide both sides by -16 to get -25/-16

= 1.56

Rounds to 1.60, or 1.6

Answer: 1.6 Seconds it will take for soccer ball to hit the ground again.

First part of the Equation: Answer: 9.8Ft. Soccer ball will reach at maximum height.

Hope that helps!!!! : )

6 0

3 years ago

A football is kicked toward an end zone with an initial vertical velocity of 30 ft/s. The function h(t) = -16+ 30t models the he

Ulleksa [173]

Answer:

A. The football does not reach a height of 15ft

Step-by-step explanation:

Given

$h(t) = -16t^2 + 30t$

Required

Determine which of the options is true

The option illustrates the height reached by the ball.

To solve this, we make use of maximum of a function

For a function f(x)

Such that:

$f(x) = ax^2 + bx + c$ :

$f(\frac{-b}{2a}) = maximum/minimum$

i.e we first solve for $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for x in $f(x) = ax^2 + bx + c$

In our case:

First we need to solve $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for t in $h(t) = -16t^2 + 30t$

By comparison:

$b = 30$

$c = -16$

$\frac{-b}{2a} = \frac{-30}{2 * -16}$

$\frac{-b}{2a} = \frac{-30}{-32}$

$\frac{-b}{2a} = \frac{30}{32}$

$\frac{-b}{2a} = \frac{15}{16}$

Substitute $\frac{15}{16}$ for t in $h(t) = -16t^2 + 30t$

$h(\frac{15}{16}) = -16(\frac{15}{16})^2 + 30(\frac{15}{16})$

$h(\frac{15}{16}) = -16(\frac{225}{256}) + \frac{450}{16}$

$h(\frac{15}{16}) = -(\frac{225}{16}) + \frac{450}{16}$

$h(\frac{15}{16}) = \frac{-225 + 450}{16}$

$h(\frac{15}{16}) = \frac{225}{16}$

$h(\frac{15}{16}) = 14.0625$

This implies that the maximum height reached is 14.0625ft.

So, the option that answers the question is A because $14.0625 < 15$

8 0

3 years ago

Find the LCM of 2 and 3

max2010maxim [7]

$\boxed{\sf \: \begin{cases}\\\begin{gathered} {{{ \sf {\red{\large\bf { {\rm {\orange{\begin{array}{r | l}2&\underline{2,3}\\3& \underline{1,3}\\&\underline{1,1}\ \: \end{array}}}}} }}}}}\end{gathered}\\\end{cases}}$

LCM = 2 × 3 = 6

LCM of 2 and 3 is 6

8 0

3 years ago

Let f(x)=√7x and g(x)=x+8, whats the smallest number that is the domain of f^o g?

Ludmilka [50]

(f○g)(x)=√(7(x+8))

(f○g)(x)=√(7x+56)

Since there is no real square root of a negative value, the smallest value in the domain would make the radicand equal to zero. x=-8 is the smallest number in the domain of (f○g)(x)

6 0

3 years ago

What is the first step needed to solve (3/4)x-3=-18 ?

inn [45]

First step is Add 3 to both sides
hope it helps

3 0

3 years ago