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Savatey [412]
3 years ago
5

the weight of an object on the moon varies directly with it weight on the earth. if an object weighing 94 Ib on the moons weight

s 570 Ib on the earth how much does an object on the moon if it weighs 2700 Ib on the earth
Mathematics
1 answer:
AlekseyPX3 years ago
7 0
446lb (rounded to the closest pound)
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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
The floor of a triangular room has an area of 32 1/2 sq.m. If the triangle’s altitude is 7 172 m, write an equation to determine
sineoko [7]

Answer:

\frac{(2)A}{h} =b

b=0.00906m

Step-by-step explanation:

Hello! To solve this exercise we must remember that the area of ​​any triangle is given by the following equation

A=\frac{bh}{2}

where

A=area=32.5m^2

h=altitude=7172m

b=base

Now what we should do take the equation for the area of ​​a rectangle and leave the base alone, remember that what we do on one side of the equation we must do on the other side to preserve equality

A=\frac{bh}{2} \\\frac{2}{h} A=\frac{bh}{2} \frac{2}{h} \\

\frac{A(2)}{h} =b

solving

\frac{2(32.5)}{7172} =0.0090[tex]\frac{A(2)}{h} =b\\b=0.00906m

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3 years ago
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Svetlanka [38]
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