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2 years ago

Im bad at composite figures || please hurry

Mathematics

1 answer:

goldfiish [28.3K]2 years ago

3 0

The figure is the combination of trapezoid and square. Then the area of the figure will be 39 square meters.

The complete question is attached below.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

The figure is the combination of trapezoid and square.

Then the area of the geometry will be

Area = Area of trapezoid + Area of square

Area = 1/2 x (3 + 9) x 5 + 3 x 3

Area = 30 + 9

Area = 39 square meters

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

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3 years ago

A)Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. If it diverges to infinity

Colt1911 [192]

Answer:

a) The sequence converges to 0

b) The lenght of the curve is $\frac{1}{54}(217^{3/2}-37^{3/2})$

Step-by-step explanation:

Consider the sequence $a_n = \frac{-6n^6 + \sin^2(7n)}{n^7+11}$

a) We will prove it using the sandwich lemma. Note that for all n $-1\leq \sin^2(7n)\leq 1$ , then

$\frac{-6n^6 -1}{n^7+11}\leq\frac{-6n^6 + \sin^2(7n)}{n^7+11}\leq \frac{-6n^6 + 1}{n^7+11}$

Note that the expressions on the left and the right hand side have a greater degree on the denominator than the one on the numerator. Then, by takint the limit n goes to infinty on both sides, we have that

$0 \leq\frac{-6n^6 + \sin^2(7n)}{n^7+11} \leq 0$

So, the sequence converges to 0.

b) The function $f(x) = 4x^{3/2}+7$ the formula of curve lenght is given by

$s = \int_a^b \sqrt[]{1+(f'(x))^2}dx$

in this case, a=1, b=6

Note that $f'(x) =6x^{{1/2}$ . Then

$s=\int_1^6 \sqrt[]{1+36x}dx$ . Take u = 1+36x. Then du= 36dx (i.e du/36 = dx). If x = 1, then u = 37 and if x = 6 then u = 217. So,

$s=\frac{1}{36}\int_{37}^{217}\sqrt[]{u} du = \frac{2}{36\cdot 3} \left.u^{3/2}\right|_{37}^{217}=\frac{1}{54}(217^{3/2}-37^{3/2})$

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