Answer: 61.58
Step-by-step explanation:
C = 2 π r = 2 · π · 9.8
π = pi
radius times pi (3.14) times 2 (because two radius makes the diameter)
9.8 times 3.14 times 2 = 61.57522in
Now we round.. 61.58
:) I hope this helps!
plz hurry!! A person standing close to the edge on top of a 108-foot building throws a ball vertically upward. The quadratic function h ( t ) = − 16 t 2 + 132 t + 108 h ( t ) = - 16 t 2 + 132 t + 108 models the ball's height above the ground, h ( t ) h ( t ) , in feet, t t seconds after it was thrown. a) What is the maximum height of the ball?
Answer:
The maximum height of the ball is 380.25 feet in the air.
Step-by-step explanation:
The quadratic function:
[tex]h(t)=-16t^2+132t+108[/tex]
Models the ball's height h(t), in feet, above the ground t seconds after it was thrown.
We want to determine the maximum height of the ball.
Note that this is a quadratic function. Therefore, the maximum or minimum value will always occur at its vertex point.
Since our leading coefficient is leading, we have a maximum point. So to find the maximum height, we will find the vertex. The vertex of a quadratic equation is given by:
[tex]\displaystyle \left(-\frac{b}{2a},f\left(\frac{b}{2a}\right)\right)[/tex]
In this case, a = -16, b = 132, and c = 108. Find the t-coordinate of the vertex:
[tex]\displaystyle t=-\frac{132}{2(-16)}=-\frac{132}{-32}=\frac{33}{8}=4.125[/tex]
So, the maximum height occurs after 4.125 seconds of the ball being thrown.
To find the maximum height, substitute this value back into the equation. Thus:
[tex]h(4.125)=-16(4.125)^2+132(4.125)+108=380.25\text{ feet}[/tex]
The maximum height of the ball is 380.25 feet in the air.
Given that the diameter of a circle is 5 mm, find the radius of the circle.
Answer:
r=2.5mm
Step-by-step explanation:
Using the formula
[tex]\huge\underline{ \sf \: Formulae \: to \: remember \: for \: solving \: the \: question:}[/tex]
[tex] \\{ \sf radius \: of \: a \: circle = \frac{d}{2} where \: d \: is \: the \: diameter \: of \: the \: circle}[/tex]
[tex]\huge\underline{\sf Given:}[/tex]
[tex] \ \: { \sf The \: diameter \: of \: a \: circle \: is \: 5 mm.}[/tex]
[tex]\huge\underline{\sf To \: find \: out:}[/tex]
[tex] \ { \sf Radius \: of \: the \: circle}[/tex]
[tex]\huge\underline{\sf Solution:}[/tex]
[tex] \\{ \sf The \: diameter \: of \: the \: circle \: is \: 5 \: mm.}[/tex]
[tex] \\{ \sf Therefore \: radius \: of \: the \: circle}[/tex]
[tex]\\{ \sf = \frac{5}{2} \: mm} [/tex]
[tex] \\{ \sf = 2.5 \: mm}[/tex]
[tex]\huge\underline{\sf Answer:}[/tex]
[tex]\\{\sf The \: radius \: of \: the \: circle \: is \: 2.5 \: mm.}[/tex]
PLEASEEEE HELPPPPP ME !! :)
Answer:
ok ok, i think it's the last one
Step-by-step explanation:
A. 34
B. 55
C. 65
D. 145