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algol13
3 years ago
10

How do u make the equation and how do u solve it

Mathematics
1 answer:
-BARSIC- [3]3 years ago
7 0

You would do the miles divided by the amount of money.

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Look at triangle ABC. What is the length of said AB of the triangle
Nuetrik [128]
\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\qquad 
B(\stackrel{x_2}{1}~,~\stackrel{y_2}{2})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AB=\sqrt{(1-4)^2+(2-5)^2}\implies AB=\sqrt{(-3)^2+(-3)^2}
\\\\\\
AB=\sqrt{9+9}\implies \boxed{AB=\sqrt{18}}\implies  AB=\sqrt{2(9)}
\\\\\\
AB=\sqrt{2(3^2)}\implies \implies AB=3\sqrt{2}
5 0
3 years ago
The amount of polonium-210 remaining, p(t), after t days in a sample can be modeled by the exponential function p(t) = 100e−0.00
Archy [21]

Answer:

% Po lost = 100[1 - e^(-0.005t)]  %; 73.0 g

Step-by-step explanation:

p(t) = 100e^(-0.005t)

Initial amount:        p(0) = 100

Amount remaining: p(t) = 100e^(-0.005t)

Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]

% of Po lost  = amount lost/initial amount × 100 %

= [1 - e^(-0.005t)]  × 100 % = 100[1 - e^(-0.005t)]  %

p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g

The mass of polonium remaining after 63 days is 73 g.

4 0
3 years ago
Triangle OMG has vertices O (6, 4), M (-8, 0), G (2, -2). It is dilated and the image coordinates are O' (3, 2) M' (-4, 0) G' (1
Harlamova29_29 [7]

Answer:

O(6,4)ML,(80),-G(2-2

6 0
3 years ago
Read 2 more answers
Point slope form of (-1,3) & (1,1)
My name is Ann [436]
-2/3

rise over run. plot the points and count.
3 0
3 years ago
What is 3/4 - (-2/3) as an improper fraction?
iragen [17]

Answer:

17/12

Step-by-step explanation:

See picture.

I wrote your poblem as an additoin problem because 2 negitive numbers would make it possitve.

Example:

1 - (-2) = 1 + 2

6 0
3 years ago
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