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2 years ago

The table represents an exponential function.

Mathematics

2 answers:

Minchanka [31]2 years ago

8 0

we know that

The multiplicative rate of change of the exponential function between two points is equal to

$[f(b) / f(a) ] / (b-a)$

Let

$A(1,6)\\B(2,4)$

we have that

$f(a)=6\\f(b)=4\\a=1\\b=2$

substitute in the formula

$[4 / 6 ] / (2-1)=4/6=2/3$

Let

$A(3,8/3)\\B(4,16/9)$

we have that

$f(a)=8/3\\f(b)=16/9\\a=3\\b=4$

substitute in the formula

$[(16/9) / (8/3) ] / (4-3)=48/72=2/3$

therefore

the answer is the option

B.) 2/3

stepladder [879]2 years ago

8 0

The best and most correct answer among the choices provided by the question is the second choice. The multiplicative rate of change of the function is 2/3. I hope my answer has come to your help. God bless and have a nice day ahead!

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Read 2 more answers

A boat sails on a bearing of 038°anf then 5km on a bearing of 067°.

I am Lyosha [343]

This question is not complete

Complete Question

A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point

Answer:

a)8.717km

b) 54.146°

Step-by-step explanation:

(a)how far is the boat from its starting point.

We solve this question using resultant vectors

= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)

Where

Rcos θ = x

Rsinθ = y

= (4cos38,4sin38) + (5cos67,5sin67)

= (3.152, 2.4626) + (1.9536, 4.6025)

= (5.1056, 7.065)

x = 5.1056

y = 7.065

Distance = √x² + y²

= √(5.1056²+ 7.065²)

= √75.98137636

= √8.7167296826

Approximately = 8.717 km

Therefore, the boat is 8.717km its starting point.

(b)calculate the bearing of the boat from its starting point.

The bearing of the boat is calculated using

tan θ = y/x

tan θ = 7.065/5.1056

θ = arc tan (7.065/5.1056)

= 54.145828196°

θ ≈ 54.146°

7 0

2 years ago

HELP!!

Liono4ka [1.6K]

Answer:

Part 1: The polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2: The vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

Step-by-step explanation:

Part 1:

The vertices of the polygon ABCDE are A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6).

The vertices of the polygon MNOPQ are M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).

We need to find the transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ.

The relation between the vertices of ABCDE and MNOPQ are defined as

$(x,y)\rightarrow (-x,y)$