2. Given: g(x) = x2 What is g(4) = ? 4 0 16 49

Suppose that P(n) is a propositional function. Determine for which improper subset of the domain of n the statement P(n) must be

Alinara [238K]

Answer:

a) P(n) is true for all 'n' in the set ; { 0,2,4,6,8 ….. }

b) P(n) is true for all 'n' in the set ; { 0,1,2,3,4,5 ............ }

Step-by-step explanation:

a) As P(0) is true

we will assume that

P(2) is true
P(4) is true
P(6) is true

this simply means that ; P(n) is true for all 'n' in the set

{ 0,2,4,6,8 ….. }

b) since P(0) and P(1) are true

we will assume that

P( 0+2 ) = P(2) is true

also P(1) and P(2) are true

we will assume that

P(1+2) = P(3) is true

Also from the previous answers it can be seen that P(2) + P(3) is true

we will assume

P(2+2) = P(4) is true

This simply means that P(n) is true for all 'n' in the set

{ 0,1,2,3,4,5 ............ }

8 0

3 years ago

Which vocabulary word best describes the math sentence you see here?

nlexa [21]

Answer:

Expression

Step-by-step explanation:

Remember, an expression is a mathematical phrase that contains numbers, variables, or both. Expressions never have an equal sign. An equation is a mathematical sentence that says two expressions are equal. My work don't copy®️

7 0

2 years ago

In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

olya-2409 [2.1K]

Answer:

$\frac{BE}{EC} =\frac{1}{3}$

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

$\frac{BK}{CK} =\frac{1}{4}$

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK and Δ EBK are similar

so we have

$\frac{AD}{BE} =\frac{DK}{BK}$

$\frac{AD}{BE} =\frac{4}{1}$

$\frac{BC}{BE}=\frac{4}{1}$ ( ABCD is parallelogram so AD=BC)

$\frac{BE+EC}{BE}=\frac{4}{1}$ ( BC= BE+EC)

$\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}$

$1+\frac{EC}{BE}=4$

$\frac{EC}{BE}=3$ ( subtracting 1 from both side )

$\frac{EC}{BE}=\frac{3}{1}$

taking reciprocal both side

$\frac{BE}{EC} =\frac{1}{3}$

8 0

4 years ago

A horse walks around a circular track while its trainer stands in the center. The trainer is 14 feet from the horse at all times

max2010maxim [7]

The horse traveled 439.6 feet after walking around the track 5 times

Solution:

Given that, horse walks around a circular track while its trainer stands in the center

The trainer is 14 feet from the horse at all times

Therefore, radius of circular track = 14 feet

The circumference of circle is the distance traveled by horse for 1 lap

The circumference of circle is given as:

$C = 2 \pi r$

Where, "r" is the radius and $\pi$ is a constant equal to 3.14

$C = 2 \times 3.14 \times 14\\\\C = 87.92$

Thus the distance traveled by horse for one time in circular track is 87.92 feet

About how far had the horse traveled after walking around the track 5 times?

Multiply the circumference by 5

$distance = 5 \times 87.92\\\\distance = 439.6$

Thus the horse traveled 439.6 feet after walking around the track 5 times

5 0

3 years ago

Use the Echelon Method 1. Solve x + 3y = 7 3x + 4y = 11

Ludmilka [50]

Answer:

x = 1 y = 2

Step-by-step explanation:

x + 3y = 7

- Subtract x from both sides.

3y = -x + 7

- Divide both sides by 3 to isolate the variable.

y = -1/3x + 7/3

- Plug the value of y into the other equation.

3x + 4(-1/3x + 7/3) = 11

3x - 4/3x + 28/3 = 11

- Add like terms.

5/3x + 28/3 = 11

5/3x = 5/3

x = 1

- Plug the value of x into the equation.

x + 3y = 7

(1) + 3y = 7

3y = 6

y = 2

5 0

3 years ago