Cell Phone Towers

HELP PLEAAAAAAAAAAASE

dalvyx [7]

The answer is option B.
Because it can't be chocolate.

4 0

3 years ago

Which of the following is the solution to |x-1|=8

Neko [114]

Waiting on the answer choices so i can answer your question :)

5 0

2 years ago

Read 2 more answers

POLYGON BASICS (GEOMETRY) PLEASE HELP!!

Vesna [10]

Answer:

Polygons are named according to the number of sides and angles they have. The most familiar polygons are the triangle, the rectangle, and the square. A regular polygon is one that has equal sides. Polygons also have diagonals, which are segments that join two vertices and are not sides.

6 0

3 years ago

Mr. Keller bought a car for $20,000. A study shows that a car will depreciate (go DOWN in value) by 15% each year. How much is M

Schach [20]

A: the formula would be f(x) = P(R) ^T or f(x) = Principle(rate)^time

B: f(x) = 20,000(0.85)^5

C: = 8,874.10625

D: Yes, the final answer makes sense compared to the origional cost of the car in relation to the formula. As well, time decreases the value of a car, so for the cost to be so low only makes sense due to the cars decrease in value or an extended and elongated amount of time.

E: You can solve this equation graphically by plotting th point at 20,000 and then taking 85% of 20,000 and plotting it each time until you get to the fifth year.

5 0

2 years ago

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Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago