Test the periodicity of the function y = tan x/4

Sharon is preparing for a trip to Kuwait. She calculates that she will need $1,166 to be converted to Kuwaiti dinar. If the

Dafna11 [192]

Answer:

A

Step-by-step explanation:

Just multiply 1166 by 0.2879

6 0

3 years ago

Change each of the following decimals to a fraction. Reduce your answer to its lowest terms.

cluponka [151]

7/100

3/10

21/125

3/ 2500

7/50

I simplified them for you

8 0

3 years ago

A quadratic equation is shown below: 3x^2 − 15x + 20 = 0 Part A: Describe the solution(s) to the equation by just determining th

Anuta_ua [19.1K]

These are two questions and two answers:

Question 1:

A quadratic equation is shown below: 3x^2 − 15x + 20 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work.

Answer: The negative value of the radicand means that the equation does not have real solutions.

Explanation:

1) With radicand the statement means the disciminant of the quadratic function.

2) The discriminant is: b² - 4ac, where a, b, and c are the coefficients of the quadratic equation: ax² + bx + c

3) Then, for 3x² - 15x + 20, a = 3, b = - 15, and c = 20

and the discriminant (radicand) is: (-15)² - 4(3)(20) = 225 - 240 = - 15.

4) The negative value of the radicand means that the equation does not have real solutions.

Question 2:

Part B: Solve 3x^2 + 5x − 8 = 0 by using an appropriate method. Show the steps of your work, and explain why you chose the method used.

Answer: two solutions x = 1 and x = - 8/3x

Explanation:

1) I choose factoring (you may use the quadratic formula if you prefer)

2) Factoring

Given: 3x² + 5x − 8 = 0

Make 5x = 8x - 3x: 3x² + 8x - 3x - 8 = 0

Group: (3x² - 3x) + (8x - 8) = 0

Common factors for each group: 3x(x -1) + 8(x - 1) = 0

Coomon factor x - 1: (x - 1) (3x + 8) = 0

The two solutions are for each factor equal to zero:

x - 1 = 0 ⇒ x = 1
3x + 8 = 0 ⇒ x = -8/3

Those are the two solutions. x = 1 and x = - 8/3

4 0

3 years ago

. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

8 0

3 years ago

Identify the term that completes the equation. AC^2 = (DC)(?) BC AD BD AB

fiasKO [112]

Answer:

BC

Step-by-step explanation:

the height of a right-angled triangle to the Hypotenuse is the square root of the product of the lengths of the 2 segments of the Hypotenuse.

so the square of the height is then the product of the 2 line segments of the Hypotenuse.

therefore

AC² = DC × BC

3 0

3 years ago