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Goryan [66]
3 years ago
5

What is the area of this figure?

Mathematics
2 answers:
VashaNatasha [74]3 years ago
8 0

Answer:

104cm^2

Step-by-step explanation:

Area of 1st rectangle

given

l=12cm

w=6cm

A =l×w

=12cm×6cm

=72cm^2

Area of 2nd rectangle

given

l=14-6=8cm

w=4cm

A=l×w

=8cm×4cm

=32cm^2

Total area =72cm^2+32cm^2=104cm^2

Bas_tet [7]3 years ago
6 0
104cm^2 is the area of that figure.
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Answer:

C.

Step-by-step explanation:

As we can see on the graph C is our answer.

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4 years ago
Solve for x. 2=4x-6​
77julia77 [94]

Answer:

x = 2

Step-by-step explanation:

Rewrite the equation as 4x - 6 = 2

Move all terms not containing x to the right side of the equation:

Add 6 to both sides of the equation.

4x = 2 + 6

Add 2 and 6.

4x = 8

Divide each term by 4 and simplify.

Divide each term in 4x = 8 by 4.

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3 years ago
Find the maclaurin series for f(x) using the definition of a maclaurin series. [assume that f has a power series expansion. do n
Nady [450]

The equation of f(x) = e^{-6x} by maclaurin series is f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i}  }{i!}.

The maclaurin series for f(x) is defined by the following formula:

f(x) = \sum_{i=0}^{\infty} \frac{f^{(i)} (0)}{i!} .x^{i}--------------(1)

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Then,

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Lastly, the equation of the trascendental function by Maclaurin series is: f(x)=\sum_{i=0}^{\infty} \frac{(-6)^{i}.x^{i}  }{i!} \\f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i}  }{i!}---------------(3)

Hence,

The equation of f(x) = e^{-6x} by maclaurin series is f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i}  }{i!}.

Find out more information about maclaurin series here

brainly.com/question/24179531

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4 0
1 year ago
luis wants to uy a skateboard that usually sells for $79.99. all merchandise is discounted by 12%. what is the total cost of the
mario62 [17]
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5 0
3 years ago
Read 2 more answers
On Novermber 27, 1993, the New York times reported that wildlife biologists have found a direct ink between the increase in the
Korolek [52]

Answer:

a.  1 263 888

b. 130 701

c. 72 years

Step-by-step explanation:

a. The differential equation applies here.

Let the quantity increase for a certain time be given by Q(t)

Every unity of time, the quantity increases by 1+\frac{r}{100} so that after the time t, the quantity remaining will be given by:

Q(t) = (1+ \frac{r}{100} )^{t}

In a similar manner, the quantity R(t) decreases at a rate given by the following expression:

1-\frac{r}{100} and after the time , t the quantity of R remaining will be given by:

R(t) = (1-\frac{r}{100} )^{t}

a. To find the population of humans in 1953

Q(t) = (1+ \frac{r}{100} )^{t}

1993 - 1953 = 40 years = t

Q(40) = Q×1.06^{40}

Q = 1 263 888.44

    ≈ 1 263 888

b. For bear population in 1993:

R(t) = (1-\frac{r}{100} )^{t}

t = 40

R(40) = b 0.94^{40} = 11 000

b = 130 700. 889

    ≈130 701

c. time taken for black bear population number less than 100 is given by:

130 = 11000×0.94^{t}

solving using natural logarithms gives t = 72.72666

                                                                  =   72 years Ans

8 0
3 years ago
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