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NNADVOKAT [17]
3 years ago
7

Determine the simple interest when:

Mathematics
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

$75.65

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 4.25%/100 = 0.0425 per year,

then, solving our equation

I = 890 × 0.0425 × 2 = 75.65

I = $ 75.65

The simple interest accumulated

on a principal of $ 890.00

at a rate of 4.25% per year

for 2 years is $ 75.65.

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marissa [1.9K]
There are  4.53592 kilograms in 10 lbs.

5 0
3 years ago
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–q + 23 + 11n + 7j + 20 – 13r<br> What are the coefficient in this expression
kobusy [5.1K]

Answer:

11,7,13

Step-by-step explanation:

coefficients come before the variable. so if it’s 13r, the variable is r. which makes the coefficient 13

4 0
3 years ago
5(x-4)+2x=14 help me find the answer x=?
Bond [772]
Exact Form:
x = 34/7

Decimal Form:
x= 4. 857142

Mixed Number Form:
x= 4 6/7
4 0
2 years ago
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Find the midpoint of the line segment with the given endpoints. 1) (-4, -9), (-6, -6)
weeeeeb [17]

Answer:

The answer is

<h2>( - 5 \: , \:  -  \frac{15}{2} )</h2>

Step-by-step explanation:

The midpoint M of two endpoints of a given line segment can be found by using the formula

<h3>M = ( \frac{x1 + x2}{2}  , \frac{y1 + y2}{2} )</h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-4, -9), (-6, -6)

The midpoint is

<h3>M = ( \frac{ - 4 - 6}{2}  , \:  \frac{ - 9 - 6}{2} ) \\  = ( -  \frac{10}{2} , -  \frac{15}{2} )</h3>

We have the final answer as

<h3>( - 5 \: , \:  -  \frac{15}{2} )</h3>

Hope this helps you

8 0
2 years ago
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio
AnnZ [28]

Answer:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

t y'' + 7y = t

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

7 0
3 years ago
Read 2 more answers
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