<h2>Answer-Average rate of change(A(x)) of f(x) over a interval [a,b] is given by:</h2><h2 /><h2>A(x) = \frac{f(b)-f(a)}{b-a}A(x)= </h2><h2>b−a</h2><h2>f(b)−f(a)</h2><h2> </h2><h2> </h2><h2 /><h2>Given the function:</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^xf(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>x</h2><h2> </h2><h2 /><h2>We have to find the average rate of change from x = 1 to x= 2</h2><h2 /><h2>At x = 1</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^1 = 5f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>1</h2><h2> =5</h2><h2 /><h2>At x = 2</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^2=20 \cdot \frac{1}{16} = 1.25f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>2</h2><h2> =20⋅ </h2><h2>16</h2><h2>1</h2><h2> </h2><h2> =1.25</h2><h2 /><h2>Substitute these in above formula we have;</h2><h2 /><h2>A(x) = \frac{f(2)-f(1)}{2-1}A(x)= </h2><h2>2−1</h2><h2>f(2)−f(1)</h2><h2> </h2><h2> </h2><h2 /><h2>⇒A(x) = \frac{1.25-5}{1}=-3.75A(x)= </h2><h2>1</h2><h2>1.25−5</h2><h2> </h2><h2> =−3.75</h2><h2 /><h2>therefore, average rate of change of the function f(x) from x = 1 to x = 2 is, -3.75</h2>
<h2>Please Mark me as brainlist. </h2>
2(3+3) or 2(3)+6, so the answer is the third option. It’s just using the distributive property.
Answer:
(-4,-4)
Step-by-step explanation:
I checked the answer in desmos
Answer:
He will save 21 dollars.
Step-by-step explanation:
44.99 - 23.99= 21
Answer:
Therefore the Correct option is First one
SAS, ∠A ≅ ∠C, AB ≅ CB , ∠ABD ≅ ∠CBD
.
Step-by-step explanation:
Given:
∠BDA ≅ ∠BDC
AD ≅ CD
TO Prove
ΔADB ≅ ΔCDB
Proof:
In ΔADB and ΔCDB
AD ≅ CD ....……….{Given}
∠BDA ≅ ∠BDC …………..{Given}
BD ≅ BD ....……….{Reflexive Property}
ΔADB ≅ ΔCDB ….{By Side-Angle-Side Congruence Postulate}
∴ ∠A ≅ ∠C ......{Corresponding Parts of Congruent Triangle are Congruent}
AB ≅ CB ......{Corresponding Parts of Congruent Triangle are Congruent}
∠ABD ≅ ∠CBD {Corresponding Parts of Congruent Triangle are Congruent}
