Supp. Newrow Support LogMeln123 Newrow Test Check Minecraft Education. Peyton Odermatt Solve: \( -5(8 x+5)+3 x=2 \) A \( x=\frac{23}{37} \) B \( x=-\frac{23}{37} \) C \( x=-\frac{27}{37} \) D \( x=\frac{27}{37} \)
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\[ -5 \cdot 8x + (-5) \cdot 5 + 3x = 2 \] \[ -40x - 25 + 3x = 2 \] Show more…
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Brief Calculus (1) Homework: Homework 3 3.2.9 Find all values x = a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist. f(x) = (x^2 - 64) / (x - 8) Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. (Use a comma to separate answers as needed.) A. f is discontinuous at the two values x = . The limit for the smaller value is . The limit for the larger value does not exist and is not inf or -inf. B. f is discontinuous at the two values x = . The limit for the smaller value is . The limit for the larger value is . C. f is discontinuous over the interval . The limit is . (Type your answer in interval notation.) D. f is continuous for all values of x. E. f is discontinuous at the two values x = . The limit for the smaller value does not exist and is not inf or -inf. The limit for the larger value is . F. f is discontinuous at the single value x = . The limit is . G. f is discontinuous at the single value x = . The limit does not exist and is not inf or -inf. H. f is discontinuous over the interval . The limit does not exist and is not inf or -inf. (Type your answer in interval notation.) I. f is discontinuous at the two values x = . The limit for both values do not exist and are not inf or -inf.
Adi S.
Need help for number 1, thank u sm.
Donna D.
If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, then (a) $\operatorname{Adj} \mathrm{A}=\mathrm{A}-4 \mathrm{I}_{3}$ (b) $\mathrm{A}^{2}=4 \mathrm{~A}-5 \mathrm{I}_{3}$ (c) $\operatorname{Adj} \mathrm{A}=\mathrm{A}+4 \mathrm{I}_{3}$ (d) $A^{2}=5 A-4 I_{3}$
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