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Elanso [62]
3 years ago
10

Write an equation to represent each function

Mathematics
1 answer:
Semenov [28]3 years ago
6 0
1:24 and 2:20 and 3:30 and 4:16 and 5:12
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Which of the following equations are dimensionally consistent?
pishuonlain [190]

Answer:

A. x = 1/2at²

Step-by-step explanation:

Among the equations, the equation that is dimensionally consistent is x=1/2at² where;

x is the distance in meters (dimension is length (L))

a is the acceleration in m/s² (dimension is L/T²)

t is the time in seconds (s) (dimension of time is T)

Substituting the dimensions into the formula to check if we are going to arrive at the dimension for distance;

1/2at² = 1/2(L/T²)T²

= 1/2× L/T² × T²

= 1/2 × L

Since L which is the resulting dimension is the dimension for distance (x), this means that the equation x = 1/2at² is dimensionally consistent.

4 0
3 years ago
Which of the summation properties or formulas require the lower index to be 1? Check all that apply.
Talja [164]

Answer:

all but the fourth option

Step-by-step explanation:

pre-calc edg 2020

3 0
3 years ago
Can anyone help me with these question
Kazeer [188]
So hmm notice the picture below

a)

the center of the circle is the midpoint of those two folks

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -10}}\quad ,&{{ -2}})\quad 
%  (c,d)
&({{ 4}}\quad ,&{{ 6}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{4-10}{2}\quad ,\quad \cfrac{6-2}{2} \right)\impliedby \textit{center of the circle}

b)

the diameter is the distance between P and Q, or the length  of that segment, and the radius is half the diameter

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -10}}\quad ,&{{ -2}})\quad 
%  (c,d)
&({{ 4}}\quad ,&{{ 6}})
\end{array}\quad 
%  distance value
d = \sqrt{({{ 4}}-{{ (-10)}})^2 + ({{ 6}}-{{ (-2)}})^2}
\\\\\\
d=\cfrac{\sqrt{(4+10)^2+(6+2)^2}}{2}\impliedby \textit{radius of the circle}

c)

so, from a), you found the h,k coordinates for the center, from b) you've got the radius

so, just plug them in here then \bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}
\end{array}


4 0
3 years ago
H=54+1/2at^2 solve for a<br> Im lost on if I should square the 1/2at first?
Verdich [7]
<h3>Answer:  a = (2h-108)/(t^2) </h3>

Other answer formats are possible.

===========================================================

Explanation:

Let's say we want to solve the equation h = 54 + 10a for the variable 'a'. This is relevant to the problem at hand don't worry.

What we'd do is two things

  1. Subtract 54 from both sides
  2. Divide by 10

So the steps look like this

h = 54 + 10a

h-54 = 10a

(h-54)/10 = a

a = (h-54)/10

Now let's say we wanted to solve this equation: h = 54 + ca, where c is a constant and can stand in place of any number. The steps would be pretty much identical. Instead of dividing both sides by 10, we divide both sides by c. We can do this as long as c isn't zero of course.

That means the equation h = 54+ca solves to a = (h-54)/c

---------------------------

Why did I do those examples? To build up to the equation your teacher gave you.

Let's replace c with 0.5t^2

This is because 1/2 = 0.5

The 1/2at^2 is the same as (1/2t^2)a

So instead of a = (h-54)/c, we'd get a = (h-54)/(0.5t^2)

As a last optional and encouraged step, we can multiply every term by 2 to get rid of that decimal value.

Doing so makes a = (h-54)/(0.5t^2) turn into a = (2h-108)/(t^2)

---------------------------

Here's another way to solve:

h = 54 + (1/2)at^2

h-54 = (1/2)at^2

2(h-54) = at^2

2h-108 = at^2

(2h-108)/(t^2) = a

a = (2h-108)/(t^2)

In the second step, I subtracted 54 from both sides. Immediately afterward, I multiplied both sides by 2 so I could clear out the fraction. Then I divided both sides by t^2 to fully isolated the variable 'a'.

6 0
3 years ago
Kai correctly used the different laws of logarithms to rewrite log27 - log(3/5) -log 5 as one logarithm.
marusya05 [52]

Answer:

A

Step-by-step explanation:

3 0
4 years ago
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