WILL GIVE BRAINLIEST! What is the constant of proportionality between cups of rice and cups of water shown in this table?

Mathematics

1 answer:

algol [13]2 years ago

7 0

Answer:

E (none of the above)

Step-by-step explanation:

1 = k(1.5)

2 = k(3)

Therefore, k = 2/3

and when we put k = 2/3 into all values in the table we notice it is right

but it could be "A" also if the relationship was cups of water to cups of rice

You might be interested in

3 4 6 9 13 18 24 What is the next number

DedPeter [7]

3 4 6 9 13 18 24 31 39 48 58
+1 +2 +3 +4 +5 +6 +7 +8 +9 +10

4 0

2 years ago

Read 2 more answers

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:

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