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3 years ago

14

What is the function graph if both the domain and range is (−∞, ∞)

1 answer:

velikii [3]3 years ago

7 0

Answer:

(−∞, ∞) and (−∞, ∞)

Step-by-step explanation:

One example (of many) might be y = 2x. The domain and range of this function are (−∞, ∞) and (−∞, ∞).

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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

oksano4ka [1.4K]

Answer:

a.) dx3x² + 2

Use the properties of integrals

That's

integral 3x² + integral 2

= 3x^2+1/3 + 2x + c

= 3x³/3 + 2x + c

= x³ + 2x + C

where C is the constant of integration

b.) x³ + 2x

Use the properties of integrals

That's

integral x³ + integral 2x

= x^3+1/4 + 2x^1+1/2

= x⁴/4 + 2x²/2 + c

= x⁴/4 + x² + C

c.) dx6x 5 + 5

Use the properties of integrals

That's

integral 6x^5 + integral 5

= 6x^5+1/6 + 5x

= 6x^6/6 + 5x

= x^6 + 5x + C

d.) x^6 + 5x

integral x^6 + integral 5x

= x^6+1/7 + 5x^1+1/2

= x^7/7 + 5/2x² + C

Hope this helps

8 0

2 years ago

____ [38]

Answers: line DB

H

D

7 0

3 years ago

2(f-9)=-28 f=?<br>-3(2t+3)=-21 t=?<br>

SVETLANKA909090 [29]

Answer:

1. f= 3/5

2. t= 3/5

Step-by-step explanation:

FYI you can use the app photo math, you just take a pic of the problem and it gives you the answer and explains the steps and it is free.

5 0

2 years ago

Read 2 more answers

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

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

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

2 years ago

How would you write "there were 3 times as many baby dinosaurs as adult dinosaurs" as an algebraic equation where x = baby dinos

Mrrafil [7]

X = 3y

hope it helped!

5 0

2 years ago

Read 2 more answers