Solve for t. t + 9 = 7t + 9 t =

Mathematics

1 answer:

belka [17]3 years ago

6 0

T would equal to 63!
7 • 9= 63

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Find the volume of each ball. use the formula. where v is the volume and r is the radius. record the volume in table A of your s

pochemuha

Answer:

The volume of tennis ball = 38.808 $cm^{3}$ .

Volume of golf ball = 33.523 $cm^{3}$ .

Step-by-step explanation:

Given: The radius of the tennis ball is 2.1cm and the radius of thr golf ball is 2.0cm. the formula is V=4πr(to the 3rd power) /3 .

To find: Volume of tennis ball and golf ball .

Formula used: Volume = $\frac{4 π r^{3} }{3}$ .

pi(π) = $\frac{22}{7}$ .

Explanation : We are given that

Radius of the tennis ball = 2.1 cm

radius of the golf ball =2.0 cm.

We need to find the volume of each ball

So , plugging the given values of radius in above formula

Volume of the tennis = $\frac{4*22(2.1)^{3} }{3*7}$

On simplification we get, volume of tennis ball = 38.808 $cm^{3}$

Volume of the golf ball= $\frac{4*22(2.0)^{3} }{3*7}$

On simplification we get, volume of golf ball = 33.523 $cm^{3}$

Therefore , The volume of tennis ball = 38.808 $cm^{3}$ .

Volume of golf ball = 33.523 $cm^{3}$ .

7 0

3 years ago

Read 2 more answers

He has 14 rows with 8 one foot squares in each row. What is the area of his peach orchard?

Inessa [10]

112 ft. You multiply 14*8 to get that, the area

3 0

3 years ago

Read 2 more answers

How do I find the Velocity and How long will it take for the ball to reach it's maximum height?

Colt1911 [192]

So hmm check the picture below

$\bf \qquad \textit{initial velocity}\\\\
h = -16t^2+v_ot+h_o \qquad \text{in feet}\\
\\ 
\begin{cases}
v_o=\textit{initial velocity of the object}\to &64\\
h_o=\textit{initial height of the object}\to &12\\
h=\textit{height of the object at "t" seconds} \end{cases}\\\\
-----------------------------\\\\$

$\bf \textit{vertex of a parabola}\\ \quad \\

\begin{array}{lccclll}
h(t)=&-16t^2&+64t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

part 1)

it takes $\bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds$

part 2)

$\bf \textit{now, doubling }v_o\\\\
\begin{cases}
v_o=\textit{initial velocity of the object}\to &128\\
h_o=\textit{initial height of the object}\to &12\\
h=\textit{height of the object at "t" seconds}\end{cases}\\\\
-----------------------------\\\\
\textit{vertex of a parabola}\\ \quad \\

\begin{array}{lccclll}
h(t)=&-16t^2&+128t&+12\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

it will reach the maximum height at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet$

how much higher than before is that? well, what was the y-coordinate for when the vₒ was 64? what did you get for $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$ ?

subtract that from this height when vₒ is 128 or doubled, to get their difference, that's how much higher it became