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Natalka [10]
3 years ago
7

Shamika's grandmother opens an account with a deposit of $5,000 for Shamika as a college savings account. The account pays 5 per

cent annual interest. How much money will be in the account after 10 years?
Mathematics
1 answer:
lesantik [10]3 years ago
7 0

Answer:

$7500

Step-by-step explanation:

$5000 was deposited

5% annual interest of the $5000 is $250

so in ten years the interest will be 10 ×$250 =$2500

so the total sum that will be in the account is $5000 + $2500 =$7500

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X=-12-6y <br>-4x+5y=-39<br>slove by the system of substitution
Artist 52 [7]
X = -6y - 12
4x + 5y =-39

                  4x + 5y = -39
     -4(-6y - 12) + 5y = -39
-4(-6y) + 4(12) + 5y = -39
       24y + 48 + 5y = -39
               29y + 48 = -39
                       29y = -87
                            y = -3

x = -6y - 12
x = -6(-3) - 12
x = 28 - 12
x = 6

(x, y) = (6, -3)
5 0
4 years ago
What is the area of the parallelogram shown below?
Maru [420]

Answer:

8

Step-by-step explanation:

8 0
3 years ago
The acceleration of an object due to gravity is 32 feet per second squared. What is acceleration due to gravity in inches per se
MrRa [10]
One foot = 12 inches.
32 feet = x inches
x/12=32/1
so 
x=12*32/1 inches per second^2
6 0
3 years ago
Read 2 more answers
a hiker begins his climb at a point a point which 214 2/5 feet below sea level . he hikes to a location at the top of the trail
kirill [66]

425 1/2 - (-214 2/5)

common denominator  10)

425 5/10 + 214 4 /10

737 9/10 ft

7 0
3 years ago
If f and g are differentiable functions for all real values of x such that f(1) = 4, g(1) = 3, f '(3) = −5, f '(1) = −4, g '(1)
belka [17]

Answer:

h'(1)=0

Step-by-step explanation:

We use the definition of the derivative of a quotient:

If h(x)=\frac{f(x)}{g(x)}, then:

h'(x)=\frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}

Since in our case we want the derivative of h(x) at the point x = 1, which is indicated by: h'(1), we need to evaluate the previous expression at x = 1, that is:

h'(1)=\frac{f'(1)*g(1)-f(1)*g'(1)}{(g(1))^2}

which, by replacing with the given numerical values:

f(1) =4\\g(1)=3\\f'(1)=-4\\g'(1)=-3

becomes:

h'(1)=\frac{f'(1)*g(1)-f(1)*g'(1)}{(g(1))^2}=\\=\frac{-4*3-4*(-3)}{(3)^2}=\frac{-12+12}{9} =\frac{0}{9} =0

3 0
3 years ago
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