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3 years ago

The sum of three consecutive numbers is 72. What is the largest of these numbers?

2 answers:

6 0

3 consecutive numbers....x, x + 1 and x + 2

x + x + 1 + x + 2 = 72
3x + 3 = 72
3x = 72 - 3
3x = 69
x = 69/3
x = 23

x + 1 = 23 + 1 = 24
x + 2 = 23 + 2 = 25

so ur numbers are 23,24,25 with 25 being the largest

Gekata [30.6K]3 years ago

3 0

<u>25</u> Is the largest number too add up to the sum.

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At 3pm, the length of the shadow of a thin vertical pole standing on level ground is the same as the height of the pole. A while

aliya0001 [1]

Answer: The height of the pole is 175.97 ( approx )

Step-by-step explanation:

Let the height of the pole is x cm,

Also, let the angle of elevation of the sun to the pole at 3 pm is $\theta$

Thus, by the question,

$tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}$

$\implies tan\theta = \frac{x}{x}$ ( at 3pm the height of shadow = height of the pole)

$\implies tan\theta = 1$

$\implies \theta = 45^{\circ}$

Again, according to the question,

When the angle of elevation is $(\theta - 12^{\circ})$ ,

The height of shadow = x + 95,

$\implies tan(45-12)^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x$

$\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}$

$\implies -0.3505924068 x = - 61.6937213538$

$x=175.969930202\approx 175.97\text{ cm}$

4 0

3 years ago

find the local and/or absolute extrema for the function over the specified domain. (Order your answers from smallest to largest

Arlecino [84]

Answer:

Minimum 8 at x=0, Maximum value: 24 at x=4

Step-by-step explanation:

Retrieving data from the original question:

$f(x)=x^{2}+8\:over\:[-1,4]$

1) Calculating the first derivative

$f'(x)=2x$

2) Now, let's work to find the critical points

Set this

$2x=0\\x=0$

0, belongs to the interval. Plug it in the original function

$f(0)=(0)^2+8\\f(0)=8$

3) Making a table x, f(x) then compare

x| f(x)

-1 | f(-1)=9

0 | f(0)=8 Minimum

4 | f(4)=24 Maximum

4) The absolute maximum value is 24 at x=4 and the absolute minimum value is 8 at x=0.

5 0

3 years ago

Help pleaseeeeeeeeeeeeeeeeeeeeeee

bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

<u>Step-by-step explanation:</u>

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$