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sdas [7]
3 years ago
7

X + 14 = 26, x = I need help please​

Mathematics
2 answers:
Lerok [7]3 years ago
5 0

Answer:

x = 12

Step-by-step explanation:

subtract 12 from the 26 and you get 14 so x = 12

xxMikexx [17]3 years ago
3 0

x=12

hope this helps :))

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The coordinates of AB are A(-5, -1) and B(-2, -6). If AC : CB = 3 : 7, what are the coordinates of point C?
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P=\left ( \frac{mx_2+nx_1}{m+n}, \ \frac{my_2+ny_1}{m+n} \right ).\\&#10;\\&#10;\text{So using this formula, we get for the given problem}\\&#10;\\&#10;C=\left ( \frac{3(-2)+7(-5)}{3+7}, \ \frac{3(-6)+7(-1)}{3+7} \right )\\&#10;\\&#10;\Rightarrow C=\left ( \frac{-6-35}{10}, \ \frac{-18-7}{10} \right )\\&#10;\\&#10;\Rightarrow C=\left ( \frac{-41}{10}, \ \frac{-25}{10} \right )\\&#10;\\&#10;\Rightarrow C=\left ( -4.1, \ -2.5 \right )

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2. Suppose 27 blackberry plants started growing in a yard. Absent constraint, the blackberry plants will spread by 80% a month.
Marta_Voda [28]

Explanation

The question indicates we should use a logistic model to estimate the number of plants after 5 months.

This can be done using the formula below;

\begin{gathered} P(t)=\frac{K}{1+Ae^{-kt}};A=\frac{K-P_{0_{}}}{P_0}_{} \\ \text{From the question} \\ P_0=\text{ Initial Plants=27} \\ K=\text{Carrying capacity =140} \end{gathered}

Workings

Step 1: We would need to get the value of A using the carrying capacity and initial plants that started growing in the yard.

This gives;

\begin{gathered} A=\frac{140-27}{27} \\ A=\frac{113}{27} \end{gathered}

Step 2: Substitute the value of A into the formula.

P(t)=\frac{140}{1+\frac{113}{27}e^{-kt}}

Step 3: Find the value of the constant k

Kindly recall that we are told that the plants increase by 80% after each month. Therefore, after one month we would have;

\begin{gathered} P(1)=27+(\frac{80}{100}\times27) \\ P(1)=\frac{243}{5} \end{gathered}

We can then have that after t= 1month

\begin{gathered} \frac{140}{1+\frac{113}{27}e^{-k\times1}}=\frac{243}{5} \\ Flip\text{ the equation} \\ \frac{1+\frac{113}{27}e^{-k}}{140}=\frac{5}{243} \\ 243(1+\frac{113}{27}e^{-k})=700 \\ 243+1017e^{-k}=700 \\ 1017e^{-k}=700-243 \\ 1017e^{-k}=457 \\ e^{-k}=\frac{457}{1017} \\ -k=\ln (\frac{457}{1017}) \end{gathered}

Step 4: Substitute -k back into the initial formula.

\begin{gathered} P(t)=\frac{140}{1+\frac{113}{27}e^{\ln (\frac{457}{1017})t}} \\ =\frac{140}{1+\frac{113}{27}(e^{\ln (\frac{457}{1017})})^t} \\ P(t)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^t} \\  \end{gathered}

The above model is can be used to find the population at any time in the future.

Therefore after 5 months, we can estimate the model to be;

\begin{gathered} P(5)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^5} \\ P(5)=\frac{140}{1.07668} \\ P(5)=130.029\approx130 \end{gathered}

Answer: The estimated number of plants after 5 months is 130 plants.

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anastassius [24]
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