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3 years ago

The sum of two numbers is 53 and the difference is 27. What are the numbers?

Mathematics

1 answer:

ikadub [295]3 years ago

8 0

Answer:

The numbers are 39 and 14.

Step-by-step explanation:

Equation 1, sum of the numbers:

x + y = 53

Equation 2, difference of the numbers:

x - y = 25

System of equations:

x + y = 53

x - y = 25

----------------

2x = 78

x = 39

Solve for y:

39 + y = 53

y = 14

The numbers are 39 and 14.

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Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

il63 [147K]

Answer:

The minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

Step-by-step explanation:

Given function is $f(x) = 0.15(x + 1)^2 - 3$

We need to find the maximum value or the minimum value for the function.

Now, differentiate $f(x) = 0.15(x + 1)^2 - 3$ w.r.t $x$ .

$f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\$

$f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)$

Now, we will equate $f'(x)=0$ to find critical point.

$0.3(x+1)=0\\x=-1$

Plug this critical point in to the function $f(x) = 0.15(x + 1)^2 - 3$ we get,

$f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3$

Also, $f''(x)=0.3$ which is positive, We have minimum value.

So, the minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

7 0

3 years ago

Carl deposited $5,142 into a savings account 20 years ago. The account has an interest rate of 3.8% and the balance is currently

Aleksandr [31]

Here is the formula you'll need
Total = Principal * (1 + (rate/n))^n*years
I don't know how to solve that for "n" so we'll use trial and error.

If compounded annually, total = 10,841.24 
If compounded quarterly, total = 10,955.64 
If compounded monthly, total = 10,981.82 
If compounded daily, total = 10,994.58 
Therefore the answer is "A", daily.

Source:
http://www.1728.org/compint3.htm

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Step-by-step explanation:

Clue: (I'm not giving the answer this is just clues) The word "absolute" helps you understand not to estimate so you want to get the RIGHT answer not the estimation.

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