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3 years ago

What will go Inside the boxes?

Mathematics

1 answer:

e-lub [12.9K]3 years ago

8 0

Answer:

18500 is the initial value

Every year they increase by a factor of 1.03

Step-by-step explanation:

This is written in the form

y = ab^x where a is the initial value and b is the growth rate/decrease

y = (18500) (1.03)^t

The initial value is 18500

The students will increase by 1.03 - 1 = .03

The factor of increase is what they multiply by which is 1.03

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If the sum of two real number is 21, what is the greatest possible value of their product

Softa [21]

Answer:

Step-by-step explanation:

let th numbers be x and y.

x+y=21

y=21-x

Product P=xy=x(21-x)=21 x-x^2

P'=21-2x

P'=0 gives 21-2x=0,x=21/2

P"=-2

which is less than 0 at x=21/2

so P is greatest at x=21/2

y=21-21/2=21/2

P=21/2 ×21/2=441/4=110.25

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3 years ago

The math score for ten of Mrs. Moore's students are shown below. How many students made a score of more than 69 but less than 90

TEA [102]

Answer:

Step-by-step explanation:

first, 70-79 have 3 students

then 80-89 have 5 students

3 0

3 years ago

Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

beks73 [17]

<h2>Answer:</h2>

The derivative of the function f(x) is:

$f'(x)=-2x$

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$

Hence, the derivative of the function f(x) is:

$f'(x)=-2x$

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Aloiza [94]

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sergejj [24]

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