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yuradex [85]
3 years ago
15

A metal rod 9.4 meters long is cut into two pieces. One piece is three times as long as the other. Find the length of the longer

piece in meters. Round your answer to the nearest 10th of a meter.
Mathematics
1 answer:
tresset_1 [31]3 years ago
7 0
2a + b = 9.4

There are 3 parts, 2 go to Rod A, 1 go to Rod B.

9.4 / 3 = 3.13333 recurring

Rod A = 2(3.13)
= 6.26

Rod B =
3.13

Rounding, Rod A = 6.3 , Rod B = 3.1
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Write as a square of a binomial: 4.8xy+36y^2+0.16 i will give brainliest plz solve
Sonja [21]

Answer:

(0.4x + 6y)^2

Step-by-step explanation:

Given

4.8xy+36y^2+0.16x^2

Required

Express as a binomial squared

4.8xy+36y^2+0.16x^2

Rewrite as:

0.16x^2+4.8xy+36y^2

Expand

0.16x^2 + 2.4xy + 2.4xy + 36y^2

Factorize:

0.4x(0.4x + 6y) + 6y(0.4x + 6y)

Factor out 0.4x + 6y

(0.4x + 6y)(0.4x + 6y)

Express as a square:

(0.4x + 6y)^2

5 0
3 years ago
Can show me how to do this ?
elena-s [515]
You have to multiply your 20cm by 6cm and your final answer being 120 because 0 times 6 is zero and 2 times 6 is 12, then that getting you 120
4 0
3 years ago
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

4 0
3 years ago
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