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Aliun [14]
3 years ago
8

Please help me I’m not good at this

Mathematics
2 answers:
gayaneshka [121]3 years ago
7 0
There's a pattern; each day she adds 9 minutes to her reading time.
54 + 9 = 63
You could also multiply the first number by 9
1*9 = 9
2*9 = 18
3*9 = 27
4*9 = 36
5*9 = 45
6*9 = 54
7*9 = 63

Hope this helps =)
Pepsi [2]3 years ago
5 0
Multiply the Day with 9

1 x 9 = 9 (True)
2 x 9 = 18
3 x 9 = 27
4 x 9 = 36
5 x 9 = 45 
6 x 9 = 54
7 x 9 = 63


63 minutes is your answer

hope this helps
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Natalie's group brought in pizza but is buying drinks at the concession stand medium sodas are a dollar and $.25 how many medium
Phantasy [73]

Keywords:

<em>Medium sodas, buy, dollars, divide </em>

For this case we must find the amount of medium sodas that Natalie's group can buy, taking into account that they have 20 dollars and that each medium soda costs 1.25 dollars. To solve, we must divide:

Let "x" be the number of medium sodas you can buy, then:

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vfiekz [6]
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On a unit circle, the vertical distance from the x-axis to a point on the perimeter of the circle is twice the horizontal distan
swat32

Check the picture below.

since the vertical distance, namely the y-coordinate, is twice as much as the horizontal, then if the horizontal is "x", the vertical one must be 2x.

let's find the hypotenuse first.

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{x}\\ b=\stackrel{opposite}{2x}\\ \end{cases} \\\\\\ c=\sqrt{x^2+(2x)^2}\implies c=\sqrt{x^2+4x^2}\implies c=\sqrt{5x^2}\implies c=x\sqrt{5} \\\\[-0.35em] ~\dotfill

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}{\stackrel{hypotenuse}{~~\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ \sqrt{5}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{\cfrac{2}{\sqrt{5}}\cdot \cfrac{\sqrt{5}}{\sqrt{5}}\implies \cfrac{2\sqrt{5}}{(\sqrt{5})^2}\implies \cfrac{2\sqrt{5}}{5}}

5 0
3 years ago
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