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2 years ago

11. Graph the function. Label the vertex and axis of symmetry. State the domain

Mathematics

1 answer:

ad-work [718]2 years ago

4 0

Answer:

x2 is the answer

Step-by-step explanation:

i found out by seeing those numbers in the numerial order.

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How many cylindrical can of bottles 12 cm high and with 6 cm diameter could be filled from a tank containing 125 liters of deter

Orlov [11]

Answer:

368 cans.

Step-by-step explanation:

Volume of 1 can = π r^2 h.

Here h (height) = 12 and r (radius) = 1/2 * 6 = 3 cm.

So V = π * 3^2 * 12

= 108π cm^3.

The tank hold 125 liters

= 125,000 cm^3, so:

Number of cans that could be filled = 125000/ 108 π

= 368.4.

4 0

2 years ago

Read 2 more answers

A right △ABC is inscribed in circle k(O, r). Find the radius of this circle if:

natta225 [31]

We have been given that a right △ABC is inscribed in circle k(O, r).

m∠C = 90°, AC = 18 cm, m∠B = 30°. We are asked to find the radius of the circle.

First of all, we will draw a diagram that represent the given scenario.

We can see from the attached file that AB is diameter of circle O and it a hypotenuse of triangle ABC.

We will use sine to find side AB.

$\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{sin}(30^{\circ})=\frac{AC}{AB}$

$\text{sin}(30^{\circ})=\frac{18}{AB}$

$AB=\frac{18}{\text{sin}(30^{\circ})}$

$AB=\frac{18}{0.5}$

$AB=36$

Wee know that radius is half the diameter, so radius of given circle would be half of the 36 that is $\frac{36}{2}=18$ .

Therefore, the radius of given circle would be 18 cm.

8 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:

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