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3 years ago

Consider the two functions given below.

Mathematics

2 answers:

g100num [7]3 years ago

5 0

Answer:

Option A is the correct answer.

Step-by-step explanation:

y intercept means x value is 0.

f(0) = 3 x 0 + 4 = 4

$g(0)=2^0-1=1-1=0$

We need same y intercepts

That is f(0) = g(0)

4 + a = 0 + b

Checking all the options

a) f(x) - 3 and g(x) + 1

4 - 3 and 0 + 1

1 and 1

Option A is correct.

b) f(x) + 4 and g(x) - 1

4 + 4 and 0 - 1

8 and -1

Option B is wrong.

c) f(x) - 2 and g(x) + 1

4 - 2 and 0 + 1

2 and 1

Option C is wrong.

d) f(x) + 3 and g(x) - 6

4 + 3 and 0 - 6

7 and -6

Option D is wrong.

So, Option A is the correct answer.

mestny [16]3 years ago

3 0

This means which transformation makes the same answer at x = 0

f(x) = 3x + 4
$g(x) = {2}^{x} - 1$

f(0) = 4
g(0) = 1 - 1 = 0

Which transformation makes 4 = 0?

4 + x = 0 + y

4 - 3 =0 + 1

1 = 1

Option 1 is your answer

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4 years ago

Read 2 more answers

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

Verify if the polygon is a square

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

The triangle BDA is a right triangle

therefore

The polygon is a square

Find the Area of the polygon

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

Find the perimeter of the polygon

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago

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Alexxandr [17]

Answer:

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Step-by-step explanation:

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vivado [14]

Answer:

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2 years ago

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Answer:

Option (b)

Step-by-step explanation:

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