All categories

3 years ago

Please help! Quick! Which table is a function?

Mathematics

1 answer:

baherus [9]3 years ago

7 0

C is the answer................

You might be interested in

What is the answer to this equation |x−1|+5=2

barxatty [35]

Answer:

No solutions

Step-by-step explanation:

Isolate the absolute value:

|x−1| + 5 = 2

Subtract 5 from both sides:

|x-1| = -3

Since an absolute value can never be equal to a negative number, there are no solutions.

6 0

2 years ago

Read 2 more answers

I don’t know how to answer my this math question, it’s easy so please help

Fofino [41]

which one ?

the bottom or the top or is this a trick

7 0

3 years ago

When you don't know the answer on a multiple-choice questions it is generally best to

cluponka [151]

When you don't know the answer on a multiple-choice questions it is generally best to guess a random answer instead of leaving it blank.
Can you follow me and like me please? I hope i can help you more.☺

5 0

3 years ago

Read 2 more answers

AHHHH please help answer and I’ll give you points

german

Answer: Yes She is correct

Step-by-step explanation:

4 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago

﻿Please help! Quick! Which table is a function?

Please help! Quick! Which table is a function?