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ivanzaharov [21]
3 years ago
14

NEED HELP QUICK!!

Mathematics
1 answer:
atroni [7]3 years ago
4 0

Answer: See explanation

Step-by-step explanation:

We are going to solve this by using the Simple interest formula which is:

= Principal × Rate × Time

= 500 × 3% × 8/12

= 500 × 0.03 × 2/3

= 10

Therefore, the interest will be $10 ane the total amount that'll be paid back will be:= $500 + $10 = $510

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Find the inverse of f(x)= (5x+1)÷ (-x+7)​
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Hope this helps :) see the attachments

6 0
4 years ago
Explain how you determine an equation for a line using a table of input/output values.
Lubov Fominskaja [6]
Example :

            x               y
            1               3
            2               6
            3               9
            4               12

first thing u do is pick any 2 points (x,y) from ur table

(1,3) and (2,6)
now we sub those into the slope formula (y2 - y1) / (x2 - x1) to find the slope
(y2 - y1) / (x2 - x1)
(1,3)....x1 = 1 and y1 = 3
(2,6)...x2 = 2 and y2 = 6
sub
slope = (6 - 3) / (2 - 1) = 3/1 = 3

now we use slope intercept formula y = mx + b
y = mx + b
slope(m) = 3
use any point off ur table...(1,3)...x = 1 and y = 3
now we sub and find b, the y int
3 = 3(1) + b
3 = 3 + b
3 - 3 = b
0 = b

so ur equation is : y = 3x + 0....which can be written as y = 3x...and if u sub any of ur points into this equation, they should make the equation true....if they dont, then it is not correct

and if u need it in standard form..
y = 3x
-3x + y = 0
3x - y = 0 ...this is standard form
6 0
3 years ago
CAN SOMEONE HELP ME WITH THIS QUESTION PLEASE?!!!
BaLLatris [955]

Answer:

Cone or Cylinder, Sphere

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the equation of a line perpendicular to y=2.5x+5
Sauron [17]

Answer:

y = -0.4 + c

Step-by-step explanation:

y = mx + c (original form)

m ⊥ - 1/m

2.5 = 25/10

25/10 ⊥ -10/25

-10/25 = -0.4

y = -0.4x + c

     

3 0
2 years ago
Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.
Jlenok [28]
y''+2y'+y=2t^2-t+3e^{-2t}

The characteristic equation is

r^2+2r+1=(r+1)^2=0\implies r=-1

so the characteristic solution is

y_c=C_1e^{-t}+C_2te^{-t}

For the particular solution, we can try looking for a solution of the form

y_p=at^2+bt+c+de^{-2t}
\implies{y_p}'=2at+b-2de^{-2t}
\implies{y_p}''=2a+4de^{-2t}

Substituting into the ODE, we have

(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}
at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}
\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3

So the general solution to the ODE is

y=y_c+y_p
y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}
4 0
3 years ago
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