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3 years ago

Calculate the value of the expression. 47 - (2 x4)

Mathematics

2 answers:

Arlecino [84]3 years ago

8 0

Answer:

Step-by-step explanation:

do 2 × 4 first to get 8 then subtract 8 from 47 to get 39

skelet666 [1.2K]3 years ago

3 0

Answer:

Step-by-step explanation:

47- (2 x 4) first multiply what's in the parentheses

47 - 8 next subtract

39 then you have your answer

47 = 39 after you get your answer you check your work

47 = 39 + 8 you just add the number you subtracted

47 = 47 if the numbers match you know you got it right

Hope this helps

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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.

8 0

1 year ago

Is -14 a natural number

Arisa [49]

Natural numbers are 1,2,3,4,5.....etc so no -14 is not a natural number

7 0

3 years ago

Read 2 more answers

Convert 2 1/5 to an<br> improper fraction.

DochEvi [55]

Answer: 11/5

Step-by-step explanation:

you would do (2) times (5) to get 10.

then you'd add 1+10 to get 11.

write this new number as the numerator... 11/5.

3 0

3 years ago

Read 2 more answers

What is the value of the interquartile range of the data below? A box-and-whisker plot. The number line goes from 0 to 52. The w

Softa [21]

Answer:

The interquartile range is 12.

Step-by-step explanation:

In a box-and-whisker plot, the left edge of the box is the lower quartile and the right edge of the box is the upper quartile. The line dividing the box is the median. The end of the left whisker shows the minimum value. The end of the right whisker shows the maximum value.

In your question, the lower quartile is 29, the upper quartile is 41, while the median is 32. The minimum value is 10 and the maximum value is 51.

The formula for the interquartile range is as follows:

Interquartile Range = Upper quartile - Lower quartile

In this question, interquartile range = 41 - 29 which equals 12. Hence, the answer is 12.

5 0

3 years ago

Read 2 more answers

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

Orlov [11]

Answer:

25.35 grams of C is formed in 14 minutes

after a long time , the limiting amount of C = 60g ,

A = 0 gram

and B = 30 grams; will remain.

Step-by-step explanation:

From the information given;

Let consider x(t) to represent the number of grams of compound C present at time (t)

It is obvious that x(0) = 0 and x(5) = 10 g;

And for x gram of C;

$\dfrac{2}{3}x$ grams of A is used ;

also $\dfrac{1}{3} x$ grams of B is used

Similarly; The amounts of A and B remaining at time (t) are;

$40 - \dfrac{2}{3}x$ and $50 - \dfrac{1}{3}x$

Therefore ; rate of formation of compound C can be said to be illustrated as ;

$\dfrac{dx}{dt }\propto (40 - \dfrac{2}{3}x)(50-\dfrac{1}{3}x)$

= $k \dfrac{2}{3}( 60-x) \dfrac{1}{3}(150-x)$

where;

k = proportionality constant.

= $\dfrac{2}{9}k (60-x)(150-x)$

By applying the separation of variable;

$\dfrac{1}{(60-x)(150-x)}dx= \dfrac{2}{9}k dt \\ \\ \\$

Solving by applying partial fraction method; we have:

$\{ \dfrac{1}{90(60-x)} - \dfrac{1}{90(150-x)} \}dx = \dfrac{2}{9}kdt$

$\dfrac{1}{90}(\dfrac{1}{x-150}-\dfrac{1}{x-60})dx =\dfrac{2}{9}kdt$

Taking the integral of both sides ; we have:

$\dfrac{1}{90}\int\limits(\dfrac{1}{x-150}- \dfrac{1}{x-60})dx= \dfrac{2}{9}\int\limits kdx$

$\dfrac{1}{90}(In(x-150)-In(x-60)) = \dfrac{2}{9}kt+C$

$\dfrac{1}{90}(In(\dfrac{x-150}{x-60})) = \dfrac{2}{9}kt+C$

$In( \dfrac{x-150}{x-60})= 20 kt + C_1 \ \ \ \ \ where \ \ C_1 = 90 C$

$\dfrac{x-150}{x-60}= Pe ^{20 kt} \ \ \ \ \ where \ \ P= e^{C_1}$

Applying the initial condition x(0) =0 to determine the value of P

Replace x= 0 and t =0 in the above equation.

$\dfrac{0-150}{0-60}= Pe ^{0}$

$\dfrac{5}{2}=P$

Thus;

$\dfrac{x-150}{x-60}=Pe^{20kt} \\ \\ \\ \dfrac{x-150}{x-60}=\dfrac{5}{2}e^{20kt} \\ \\ \\ 2x -300 =5e^{20kt}(x-60)$

$2x - 300 = 5xe^{20kt} - 300 e^{20kt} \\ \\ 5xe^{20kt} -2x = 300 e^{20kt} -300 \\ \\ x(5e^{20kt} -2) = 300 e^{20kt} -300 \\ \\ x= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$