Which is the smallest ratio 3 to 4 2 / 3 10 :2 ,2 to 1

This is my last question please help asap!!

Ivanshal [37]

The average rate of change of function $g(x)=5(2)^{x}$ from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

The correct option is (A).

What is the average rate of change of a function?

The average rate at which one quantity changes in relation to another's change is referred to as the average rate of change function.

Using function notation, we can define the Average Rate of Change of a function f from a to b as:

$rate(m) = \frac{f(b)-f(a)}{b-a}$

The given function is $g(x) = 5(2)^{x}$ ,

Now calculating the average rate of change of function from x = 1 to x = 2.

$m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(2)-g(1)}{2-1}\\\\m=\frac{5(2)^{2}-5(2)^1}{2-1}\\ m=\frac{10}{1} \\m=10$

Now, calculate the average rate of change of function from x = 3 to x = 4.

$m = \frac{g(b)-g(a)}{b-a}\\ m = \frac{g(4)-g(3)}{4-3}\\\\m=\frac{5(2)^{4}-5(2)^3}{4-3}\\ m=\frac{40}{1} \\m=40$

The jump from m = 10 to m = 40 is "times 4".

So option (A) is correct.

Hence, The average rate of change of function $g(x)=5(2)^{x}$ from x = 3 to x = 4 is 4 times that from x = 1 to x = 2.

To learn more about the average rate of change of function, visit:

brainly.com/question/24313700

#SPJ1

7 0

1 year ago

Derivative of <br> <img src="https://tex.z-dn.net/?f=x%5E%7B3%7D" id="TexFormula1" title="x^{3}" alt="x^{3}" align="absmiddle" c

ch4aika [34]

$f(x)=x^3e^{2x}\\\\f'(x)=(x^3e^{2x})'\qquad\text{use}\ (f\cdot g)'(x)=f'(x)g(x)+f(x)g'(x)\\\\=(x^3)'(e^{2x})+(x^3)(e^{2x})'\\\\\text{use}\ (x^n)'=nx^{n-1}\ \text{and}\ (e^x)'=e^x\ \text{and}\ f'[g(x)]=f'(g(x))\cdot g'(x)\\\\=3x^2e^{2x}+x^3e^{2x}\cdot2\\\\=\boxed{3x^2e^{2x}+2x^3e^{2x}}$

5 0

3 years ago

I need help with this proof

Maksim231197 [3]

Answer:

See explanation

Step-by-step explanation:

Consider two triangles ABF and EDF. These triangles are two right triangles, because angles B and D are right anges.

In these triangles,

∠ABF ≅ ∠EDF - as two right angles;
BF ≅ FE - given;
∠AFB ≅ ∠EFD - as vertical angles when two lines AD and BE itersect.

By ASA postulate (or HA postulate) these triangles are congruent, so

ΔABF ≅ ΔEDF

Congruent triangles have congruent corresponding parts, so

FA ≅ EF

6 0

3 years ago

Help please :c!

LUCKY_DIMON [66]

Answer:

see the procedure

Step-by-step explanation:

we know that

m∠1+m∠2=180° -----> supplementary angles (form a linear pair)

we have

m∠1=27° ----> given problem

substitute the measure of m∠1 in the equation above and solve for m∠2

27°+m∠2=180°

Subtract 27° both sides

27°+m∠2-27°=180°-27° ----> by subtraction property of equality

m∠2=153°

4 0

3 years ago

150 calories and 30 grams how many calories are in 7 grams

exis [7]

Answer:

35 calories

Step-by-step explanation:

create a proportion and solve.

30/150= 7/x

7 0

3 years ago