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3 years ago

How do you change 2.3 - 3.94 into an addition problem?

Mathematics

2 answers:

Elena L [17]3 years ago

8 0

Answer:

2.3+ (-3.94)

Step-by-step explanation:

Nonamiya [84]3 years ago

7 0

Answer:

Solve it then add the answer and one of the parts of the original.

Step-by-step explanation:

Ex.

Original: 2.3 - 3.94

2.3 - 3.94 = -1.64

New addition: -1.64 + 3.94 = 2.3

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The Brooklyn Botanical Garden is 25 miles away from the Bronx Zoo. On a map of the city, they are shown as 5 inches

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Answer:

5 Miles apart

Step-by-step explanation:

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3 years ago

Se lanza un objeto desde una plataforma.

jasenka [17]

Answer:

6 seconds

Step-by-step explanation:

<u><em>The question in English is</em></u>

An object is thrown from a platform.

Its height (in meters), x seconds after launch, is modeled by:

h(x)=-5x^2+20x+60

How many seconds after the launch does the object reach the ground?

Let

x ----> the time in seconds

h(x) ---> the height of the object

we have

$h(x)=-5x^2+20x+60$

we know that

When the object hit the ground the height is equal to zero

For h(x)=0

we have

$-5x^2+20x+60=60$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-5x^2+20x+60=60$

$a=-5\\b=20\\c=60$

substitute in the formula

$x=\frac{-20\pm\sqrt{20^{2}-4(-5)(60)}} {2(-5)}$

$x=\frac{-20\pm\sqrt{1,600}} {-10}$

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The solution is x=6 sec

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Step-by-step explanation:

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$\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{0}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{162}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+0t+162\implies 0=-16t^2+162\implies 16t^2=162
\\\\\\
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3 years ago

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antiseptic1488 [7]

Answer:

Step-by-step explanation:

Given

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4 years ago

Read 2 more answers