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1 year ago

11

Have someone who can can help me with this I need help

2 answers:

denpristay [2]1 year ago

6 0

BD: 16.8+50.4= 67.2 mm
LM: 4-2.6= 1.4 cm

yan [13]1 year ago

3 0

Answer is underlined

BD: 16.8+50.4= <u>67.2 mm</u>

LM: 4-2.6= <u>1.4 cm</u>

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Find S5 for a geometric series for which a1=81 and r=1/9.

Minchanka [31]

ANSWER

$S_5=91\frac{10}{81}$

EXPLANATION

The sum of the first $n$ terms of a geometric sequence is given by;

$S_n=\frac{a_1(1-r^n)}{1-r} ,-1$

Where $n$ , is the number of terms and $a_1$ is the first term.

When $n=5$ , we have $a_1=81$ , we get;

$S_5=\frac{81(1-(\frac{1}{9})^5)}{1-\frac{1}{9}}$

$S_5=\frac{81(1-\frac{1}{59049})}{1-\frac{1}{9}}$

$S_5=\frac{81(\frac{59048}{59049})}{\frac{8}{9}}$

$S_5=\frac{7381}{81}$

$S_5=91\frac{10}{81}$

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3 years ago

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3 years ago

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3 years ago

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3 0

2 years ago

Read 2 more answers

Evaluate using <br> Definite integrals

swat32

Since $[0,4]=[0,1]\cup(1,4]$ , we can rewrite the integral as

$\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt$

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

$\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt$

Both integrals are quite immediate: you only need to use the power rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}$

to get

$\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4$

Now we only need to evaluate the antiderivatives:

$\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15$

So, the final answer is 15.

4 0

3 years ago