Janet has one rectangular-shaped paper with a length of 11/12 foot and a width of 3/4 foot.

Mathematics

1 answer:

Mice21 [21]3 years ago

5 0

Hi bestie lol i’m failing math too and use this app as a lil cheat :)

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What is the percent increased from 22.5 to 27.5

Flauer [41]

Answer:6.27

Step-by-step explanation: 22.8 increased by 27.5%

27.5% × 22.8 = 6.27

27.5 ÷ 100 × 22.8 = 6.27

27.5 × 22.8 ÷ 100 = 6.27

627 ÷ 100 = 6.27

4 0

3 years ago

In a geometric sequence, the ratio between consecutive terms is...

Leokris [45]

<h3>Answer: The same or equal</h3>

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

Explanation:

Consider an example like 3, 6, 12, 24, 48, ...

The ratio between consecutive terms is

6/3 = 2
12/6 = 2
24/12 = 2
48/2 = 2

Each time we divide any given term over its previous one, we get the same ratio 2. We call this the common ratio. In terms of notation, the variable r is used for the common ratio, so r = 2 in this case.

As another example, the geometric sequence 5, 50, 500, 5000, ... has r = 10 as the common ratio because we multiply each term by 10 to get the next one. Moving forward has us multiply by r, moving backward and we divide by r. The value of r cannot be zero, but it can be negative.

An example with r being negative would be something like

1, -1, 1, -1, 1, -1, ....

we just bounce back and forth between those two values. In this case, r = -1.

6 0

3 years ago

Suppose a punter kicks a football so that the upward component of its velocity is 80 feet per second. If the ball is 3 feet off

Alexxandr [17]

So hmm check the picture below

$\bf \qquad \textit{initial velocity}\\\\ h = -16t^2+v_ot+h_o \qquad \text{in feet}\\
\\
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}\\\\
-----------------------------\\\\
h(t)=3+80t-16t^2\iff h(t)=-16t^2+80t+3$

a)

well, clearly is 80 ft/s

b)

when t = 1? well 80(1)

c)

in the picture, x-axis is the time and y-axis is the height
so, it reaches its maximum at the vertex, after "x" seconds

$\bf \begin{array}{lcccll}
h(t)=&-16t^2&+80t&+3\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so it reached the vertex after $\bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds$

d)

the maximum height of the ball is $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet$