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2 years ago

How do i explain 5×23×2

Mathematics

1 answer:

Zarrin [17]2 years ago

6 0

Answer:

Step-by-step explanation:

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AleksAgata [21]

$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\textit{sphere A}}{\textit{sphere B}}\qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}\qquad \qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}=\stackrel{\stackrel{volumes'}{ratio}}{\cfrac{\sqrt[3]{288}}{\sqrt[3]{v}}}\implies \cfrac{1}{4}=\sqrt[3]{\cfrac{288}{v}}\implies \left( \cfrac{1}{4} \right)^3=\cfrac{288}{v} \\\\\\ \cfrac{1^3}{4^3}=\cfrac{288}{v}\implies \cfrac{1}{64}=\cfrac{288}{v}\implies v=18432$

3 0

3 years ago

Alan can type 80 words in two

Sonbull [250]

Answer:

2400 words

Step-by-step explanation:

80 = 2 min

60min ÷ 2min = 30min

30 × 80 = 2400 words

3 0

2 years ago

Read 2 more answers

What is the equation of the line that represents the horizontal asymptote of the function f(x)=25,000(1+0.025)^(x)?

posledela

Answer:

The answer is below

Step-by-step explanation:

The horizontal asymptote of a function f(x) is gotten by finding the limit as x ⇒ ∞ or x ⇒ -∞. If the limit gives you a finite value, then your asymptote is at that point.

$\lim_{x \to \infty} f(x)=A\\\\or\\\\ \lim_{x \to -\infty} f(x)=A\\\\where\ A\ is\ a\ finite\ value.\\\\Given\ that \ f(x) =25000(1+0.025)^x\\\\ \lim_{x \to \infty} f(x)= \lim_{x \to \infty} [25000(1+0.025)^x]= \lim_{x \to \infty} [25000(1.025)^x]\\=25000 \lim_{x \to \infty} [(1.025)^x]=25000(\infty)=\infty\\\\ \lim_{x \to -\infty} f(x)= \lim_{x \to -\infty} [25000(1+0.025)^x]= \lim_{x \to -\infty} [25000(1.025)^x]\\=25000 \lim_{x \to -\infty} [(1.025)^x]=25000(0)=0\\\\$

$Since\ \lim_{x \to -\infty} f(x)=0\ is\ a\ finite\ value,hence:\\\\Hence\ the\ horizontal\ asymtotes\ is\ at\ y=0$

5 0

2 years ago

Triangle NMO has vertices at N(−5, 2), M(−2, 1), O(−3 , 3). Determine the vertices of image N′M′O′ if the preimage is reflected

Firdavs [7]

Notice that the reflection is over a line of the form x=constant; in this case, the y-coordinate of the reflected point stays the same while the x-coordinate changes as expressed by the transformation below

$\begin{gathered} y\rightarrow y \\ x\rightarrow2a+x \\ where \\ x=a\rightarrow\text{ vertical line} \end{gathered}$

Hence, in our case

$\Rightarrow(x,y)=(2*-4-x,y)=(-8-x,y)$

Transform points N, M, and O accordingly,

$\begin{gathered} \Rightarrow N^{\prime}=(-8-(-5),2)=(-8+5,2)=(-3,2) \\ \Rightarrow M^{\prime}=(-8-(-2),1)=(-6,1) \\ \Rightarrow O^{\prime}=(-8-(-3),3)=(-5,3) \end{gathered}$ <h2>Therefore, the answer is the first option (top to bottom)</h2>

6 0

6 months ago

⚠️‼️ HELP PLEASE GIVING POINTS !!

vitfil [10]

Answer:

m∠QPR = 25°.

Step-by-step explanation:

If m∠QPR = 9x + 16°, substitute in 1 for 'x':. 9(1) + 16 = 9 + 16 = 25°

7 0

1 year ago