Edhesive 3.2 code practice question 3

I don't have the text of Edhesive 3.2 Question 3 — please paste the exact problem (and any code you've tried) so I can solve and explain it. What I need from you The full problem prompt (word-for-word if possible) The language (usually Java for Edhesive AP CS A) Any code you wrote / attempted Any error messages or failing test cases What I'll provide after you paste it A direct, correct solution (code) labeled clearly. Step-by-step explanation of the logic and how the code works. Common edge cases and test examples. Suggestions to improve style or efficiency if relevant. If you want, paste your code now and I’ll debug it line-by-line and explain the fix.

Which formula/name pair is incorrect? A) FeSO4 iron(II) sulfate B) Fe2 (SO3)3 iron(III) sulfite C) FeS iron(II) sulfide D) FeSO3 iron(II) sulfite E) Fe2 (SO4)3 iron(III) sulfide

The correct answer is E. Explanation Fe2(SO4)3 contains sulfate ions $SO_4^{2-}$. Three sulfates give $3\times(-2)=-6$, so two iron atoms must provide $+6$ total, i.e. $Fe^{3+}$. Therefore Fe2(SO4)3 is iron(III) sulfate, not iron(III) sulfide. Iron(III) sulfide (with $S^{2-}$) would have formula Fe2S3. The other pairs are correct: FeSO4 = iron(II) sulfate, Fe2(SO3)3 = iron(III) sulfite, FeS = iron(II) sulfide, FeSO3 = iron(II) sulfite.

(i) How many patterns does he need to form the smallest square? (ii) What are the dimensions of the largest square that he can form?

I can’t give the numeric answers yet — I need the full problem statement (how many patterns he has and the size/shape of each pattern). Explanation / what I need Please tell me whether the "patterns" are identical tiles, and give their dimensions (or area), and whether the supply is limited (how many patterns he actually has). Without that, there are several common interpretations; below I show the standard methods so you can apply them once you supply the missing data. If the question is: "How many identical rectangular patterns (each of size a×b) are needed to make the smallest square?" Let the area of one pattern be $m=a\\times b$. You need the smallest integer $N$ such that $N\\cdot m$ is a perfect square. Factor $m=\\prod p_i^{e_i}$. For each prime $p_i$ with odd exponent $e_i$, include one extra $p_i$ in $N$ to make the exponent even. Equivalently, write $m=u^2\\cdot t$ where $t$ is squarefree; then the minimal $N$ is $t$. Example: if a pattern is $2\\times3$ so $m=6=2\\cdot3$, then $t=6$ and the smallest $N$ is $6$ (since $6\\cdot6=36$, a perfect square). The resulting square will have area $36$ and side length $6$ (if units are unit squares). If the question is: "Given a finite number $T$ of identical patterns, what is the largest square (by number of patterns) he can form?" The largest number of whole patterns you can arrange into a perfect square count is the largest perfect square ≤ $T$, i.e. $\lfloor\\sqrt{T}\\rfloor^2$ patterns. The side length (in patterns) is $\lfloor\\sqrt{T}\\rfloor$. Example: if he has $T=20$ patterns, the largest square uses $\\lfloor\\sqrt{20}\\rfloor^2=4^2=16$ patterns, arranged as a $4\\times4$ square of patterns. If you paste the exact problem (pattern dimensions and how many he has), I’ll compute (i) the minimal number needed for the smallest square and (ii) the dimensions of the largest square he can form, with full step-by-step working.

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Here is a different number machine: 13 Input: 17 When the input is 17, the output is 10. Complete the number machine. Output: ______

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Here is a different number machine: 13 Input: 17 When the input is 17, the output is 10. Complete the number machine. Output: ______

Q: Here is a different number machine: 13 Input: 17 When the input is 17, the output is 10. Complete the number machine. Output: ______

The correct answer is: $10$ Explanation From the example Input 17 → Output 10 we deduce the rule is: subtract the product of the digits from the number. Steps: Product of digits of 17: $1\times7=7$ Subtract from 17: $17-7=10$ Apply same rule to 13: product $1\times3=3$, then $13-3=10$ Therefore the output is $10$.

Q: What equation is represented by this model, and what is the value of c?

I can’t determine the equation or the value of c from your message — please upload the model image or paste the equation or the data points it uses. Explanation and how I can help once you provide the model If the model is a quadratic in standard form $y=ax^2+bx+c$, then $c$ is the y‑intercept (the value of $y$ when $x=0$). So plug in $x=0$ or read the graph at $x=0$ to get $c$. If the model is a quadratic in vertex form $y=a(x-h)^2+k$, expand or evaluate at $x=0$ to find $c$: $c=a(0-h)^2+k$. If the model is exponential $y=ab^x+c$, then $c$ is the horizontal asymptote (the value $y$ approaches as $x\to\infty$ if $0<b<1$, or as $x\to-\infty$ if $b>1$). You can read the asymptote from the graph or solve for $c$ after finding $a$ and $b$ from data points. If the model is linear $y=mx+c$, then again $c$ is the y‑intercept (value at $x=0$). For other models (logistic, sinusoidal, etc.) tell me the form and I’ll show how to identify the parameter named $c$. Give me the model image, the equation, or two/three labeled points from the graph and I’ll compute the equation and the value of $c$ step by step.

Q: What is the angular velocity of the satellite in radians per minute? How far does the satellite travel in 1 hour?

I can’t give a numeric answer without the satellite’s orbital period or its orbital radius/altitude — please provide the orbital period $T$ (in minutes) or the orbital radius $r$ (in km or m). Explanation Angular speed for a circular orbit (in radians per minute) is $$\omega=\frac{2\pi}{T}$$where $T$ is the orbital period in minutes. Distance traveled in time $t$ (e.g., $t=60\text{ min}$ for 1 hour) along a circular orbit of radius $r$ is $$s=r\theta=r\omega t=r\left(\frac{2\pi}{T}\right)t.$$Equivalently, the fraction of an orbit covered in $t$ minutes is $t/T$, so$$s=\frac{t}{T}\cdot 2\pi r.$$ Example (typical low Earth orbit): if $T=90\text{ min}$ and altitude $\approx400\text{ km}$, then $r\approx6371+400=6771\text{ km}$. $\omega=\dfrac{2\pi}{90}\approx0.0698\ \text{rad/min}.$ In one hour ($t=60\text{ min}$), fraction of orbit $=60/90=2/3$, so $$s=\frac{60}{90}\cdot2\pi\cdot6771\text{ km}\approx2.836\times10^4\text{ km}.$$ If you give me the actual $T$ or $r$ (or altitude), I’ll compute the exact $\omega$ and the distance for 1 hour.