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Gelneren [198K]
2 years ago
8

It takes a bear 9 more days to eat a barrel of honey alone than it takes the bear and a second bear to eat the barrel of honey t

ogether. It takes the first bear 7 fewer days to eat a barrel of honey alone than it takes the second bear to eat a barrel of honey alone. In how many days can the two bears eat a barrel of honey together?
No links please.
Mathematics
2 answers:
algol132 years ago
7 0

Answer:

it takes the bears 12 days to eat honey together.

Step-by-step explanation:

1/x+9+1/x+16=1/x

(solve this using cross multiplication)

= x^2 = 144

x=12

romanna [79]2 years ago
4 0

answer: 12

explanation:

let x=two bears eating honey together

let x+9=first bear

let (x+9)+7=second bear     **because we know the second day takes 7 more days to eat the same amount of honey as the first bear**

since the bears can finish a barrel of honey over multiple days, they can only finish part of the barrel of honey in one day.

therefore...

x/(x+9)+x(x+16)=1 or x*1/(x+9)+x*1/(x+16)=1     **the equation equals 1 because you are trying to find how much honey they eat in one day aka 1=single day**

if you solve, you get x=-12; x=12. it can only be the positive value so x=12 is your answer. your welcome fellow rsm students.

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log (PQ) = log P + log Q

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Step-by-step explanation:

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mafiozo [28]

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8-5 + 4x> 5x\\3 + 4x> 5x

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If d = 2, what is the value of ?
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Two cities are 45 miles apart. Two trains, with speeds of 70 mph and 60 mph, leave the two cities at the same time so that one i
KATRIN_1 [288]

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4 0
3 years ago
Read 2 more answers
Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Co
Volgvan

Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like v_{1},v_{2},v_{3}, ... we call them linearly dependent if we have scalars a_{1},a_{2},a_{3},... as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0  

When all scalar coefficients are equal to zero, we can call them linearly independent

2)  Now let's examine the Matrix given:

\begin{bmatrix}1 &-2  &2  &3 \\ -2 & 4 & -4 &3 \\ 0&1  &-1  & 4\end{bmatrix}

So each column of this Matrix is a vector. So we can write them as:

\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle Or

Now let's rewrite it as a system of equations:

a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

 \left ( \left.\begin{matrix}1 &-2  &2  &3 \\ -2 &4  &-4  &3 \\ 0 & 1 &-1  &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )R_{1}\times2 +R_{2}\rightarrow R_{2}\left ( \left.\begin{matrix}1 &-2  &2  &3 \\ 0 &0 &9  &0\\ 0 & 1 &-1  &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\ R_{2}\Leftrightarrow  R_{3}\left ( \left.\begin{matrix}1 &-2  &2  &3 \\ 0 &1  &-1  &4 \\ 0 &0 &9  &0 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\left\{\begin{matrix}1a_{1} &-2a_{2}  &+2a_{3}  &+3a_{4}  &=0 \\  &1a_{2}  &-1a_{3} &+4a_{4}  &=0 \\  &  &  &9a_{4}  &=0 \end{matrix}\right.\Rightarrow a_{1}=0, a_{2}=a_{3},a_{4}=0

S=\begin{bmatrix}0\\ a_{3}\\ a_{3}\\ 0\end{bmatrix}

3 0
3 years ago
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