Write a general formula to describe the variation: x varies jointly with the inverse of the sum of y and z.

A = {a, b, c, d}

Helen [10]

Answer:

Johnny is wrong.

$A \cup B = \left\{a,b,c,d,e,f,g,h\right\}$

Step-by-step explanation:

Johnny is wrong.

A better definition would be: $A \cup B$ is the set of all elements that belong to at least one A or B.

So, the elements that belong to both A and B, like c and d in this exercise, also belong to $A \cup B$ .

So:

$A \cup B=\left\{a,b,c,d,e,f,g,h\right\}$

4 0

2 years ago

The length of a rectangle is 6 inches more than its width. Its area is 91 square inches. The length is .

Vika [28.1K]

Answer:

13 in

Step-by-step explanation:

let w be width then length is w + 6

the area (A) of a rectangle is calculated as

A = length × width , then

A = w(w + 6) = 91 , that is

w² + 6w = 91 ( subtract 91 from both sides )

w² + 6w - 91 = 0 ← in standard quadratic form

(w + 13)(w - 7) = 0 ← in factored form

equate each factor to zero and solve for w

w + 13 = 0 ⇒ w = - 13

w - 7 = 0 ⇒ w = 7

however, w > 0 then w = 7

and length = w + 6 = 7 + 6 = 13 in

8 0

1 year ago

Please help!

kotegsom [21]

Answer:

7

Step-by-step explanation:

f(x) means y, it is equal to the value of y, so f(x) = y

so, f(1) means to find the value of y when the value of x is 1, and that is 7, since 7 is the value of y indicated when the value of x is 1.

I hope this is helpful.

4 0

2 years ago

The fourth term in an arithmetic sequence is 18 and the seventh term is 39. If the first term is a1 , which is an equation for t

evablogger [386]

Answer:

Step-by-step explanation:

The nth term: $a_n=a_1+d(n-1)$

So:

a₄ = a₁ + 3d

18 = a₁ + 3d ⇒ 3d = 18 - a₁

a₇ = a₁ + 6d

39 = a₁ + 6d

39 = a₁ + 2(18 - a₁)

39 = a₁ + 36 - 2a₁

39 = 36 - a₁

a₁ = -3

3d = 18 - (-3)

3d = 21

d = 7

So the nth term:

$a_n = -3 + 7(n-1)\\\\a_n=-3 +7n - 7\\\\a_n = 7n - 10$

8 0

2 years ago

Can you please answer the question?

Roman55 [17]

The <em>trigonometric</em> expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the <em>trigonometric</em> expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that <em>one</em> side of the equality is equal to the expression of the <em>other</em> side, requiring the use of <em>algebraic</em> and <em>trigonometric</em> properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The <em>trigonometric</em> expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the <em>trigonometric</em> expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago