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3 years ago

Find the 90th term of the arithmetic sequence 16 , 21 , 26

Mathematics

1 answer:

Artist 52 [7]3 years ago

6 0

Answer:

The 90th term of the arithmetic sequence is 461.

Step-by-step explanation:

Arithmetic sequences concepts:

The general rule of an arithmetic sequence is the following:

$a_{n+1} = a_{n} + d$

In which d is the common diference between each term.

We can expand the general equation to find the nth term from the first, by the following equation:

$a_{n} = a_{1} + (n-1)*d$

In this question:

$a_{1} = 16, d = 21 - 16 = 26 - 21 = 5$

90th term

$a_{90} = a_{1} + (90-1)*d$

$a_{90} = 16 + 89*5$

$a_{90} = 461$

The 90th term of the arithmetic sequence is 461.

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3 years ago

What is 2logx−logy+2logz written as a single logarithm? log(xz)^2/y log2x/2yz logx^2y/z^2 logx^2/yz^2

Klio2033 [76]

Answer: $\log\dfrac{x^2z^2}{y}$

Step-by-step explanation:

Properties of logarithm:

$(i)\ \ n\log a = a^n\\\\ (ii)\ \ \log m +\log n =\log (mn)\\\\ (iii)\ \ \log m-\log n =\log\dfrac{m}{n}$

Consider,

$2\log x-\log y+2 \log z\\\\ =\log x^2-\log y+\log z^2\ \ \ \ \text{[By (i)]}\\\\= \log x^2+\log z^2-\log y\\\\=\log(x^2z^2)-\log y\ \ \ \ [\text{By } (ii) ]\\\\=\log(\dfrac{x^2z^2}{y}) \ \ \ \ [\text{By (iii)}]$

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3 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

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3 years ago