Radical form of f(x)=3+2 in X

3(5-x)<5x-17<br><br><br>x>-4<br>x<-4<br>x<4<br>x>4

solmaris [256]

Answer: x > 4

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7 0

3 years ago

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What is the average rate of change of the functionf(x)=20(14)x from x = 1 to x = 2?

atroni [7]

<h2>Answer-Average rate of change(A(x)) of f(x) over a interval [a,b] is given by:</h2><h2 /><h2>A(x) = \frac{f(b)-f(a)}{b-a}A(x)= </h2><h2>b−a</h2><h2>f(b)−f(a)</h2><h2> </h2><h2> </h2><h2 /><h2>Given the function:</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^xf(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>x</h2><h2> </h2><h2 /><h2>We have to find the average rate of change from x = 1 to x= 2</h2><h2 /><h2>At x = 1</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^1 = 5f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>1</h2><h2> =5</h2><h2 /><h2>At x = 2</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^2=20 \cdot \frac{1}{16} = 1.25f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>2</h2><h2> =20⋅ </h2><h2>16</h2><h2>1</h2><h2> </h2><h2> =1.25</h2><h2 /><h2>Substitute these in above formula we have;</h2><h2 /><h2>A(x) = \frac{f(2)-f(1)}{2-1}A(x)= </h2><h2>2−1</h2><h2>f(2)−f(1)</h2><h2> </h2><h2> </h2><h2 /><h2>⇒A(x) = \frac{1.25-5}{1}=-3.75A(x)= </h2><h2>1</h2><h2>1.25−5</h2><h2> </h2><h2> =−3.75</h2><h2 /><h2>therefore, average rate of change of the function f(x) from x = 1 to x = 2 is, -3.75</h2>

<h2>Please Mark me as brainlist. </h2>

8 0

2 years ago

A={ 0,1,3,5,6}; B={1,2,3,7} draw the venn diagram of AUB Writ

Sati [7]

Answer:

Here,

n(A)={0,1,3,5,6}

n(B)={1,2,3,7}

n(AUB)=?

we know that,

AUB={0,1,2,3,5,6,7}

Step-by-step explanation:

is it okay ?

hopefully it works <3 ❣️

8 0

2 years ago

In quadrilateral ABDC, AB ∥ CD. Which additional piece of information is needed to determine that ABDC is a parallelogram?

N76 [4]

Since the definition of a parallelogram literally says that it's a quadrilateral with 2 parallel sides, we have AB || CD so the 2 other sides (AC and BD) need to be either similar or parallel (either works, due to that if they're similar we can prove that since AB || CD that they're parallel). B is our answer

3 0

3 years ago

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He vertices of square pqrs are p -4,0 q 4,3 r 7,-5 and s -1,-18.Show that the diagonals of square pqrs are congruent perpendicul

Anit [1.1K]

Answer:

Step-by-step explanation:

The vertices of the square given are P(-4, 0), Q(4, 3), R(7, -5) and, S(-1, -18)

For this diagonal to be right angle the slope of the diagonal must be m1=-1/m2

So let find the slope of diagonal 1

The two points are P and R

P(-4, 0), R(7, -5)

Slope is given as

m1=∆y/∆x

m1=(y2-y1)/(x2-x1)

m1=-5-0/7--4

m1=-5/7+4

m1=-5/11

Slope of the second diagonal

Which is Q and S

Q(4, 3), S(-1, -18)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-18-3)/(-1-4)

m2=-21/-5

m2=21/5

So, slope of diagonal 1 is not equal to slope two

This shows that the diagonal of the square are not diagonal.

But the diagonal of a square should be perpendicular, this shows that this is not a square, let prove that with distance between two points

Given two points

(x1,y1) and (x2,y2)

Distance between the two points is

D=√(y2-y1)²+(x2-x1)²

For line PQ

P(-4, 0), Q(4, 3)

PQ=√(3-0)²+(4--4)²

PQ=√(3)²+(4+4)²

PQ=√9+8²

PQ=√9+64

PQ=√73

Also let fine RS

R(7, -5) and, S(-1, -18)

RS=√(-18--5)+(-1-7)

RS=√(-18+5)²+(-1-7)²

RS=√(-13)²+(-8)²

RS=√169+64

RS=√233

Since RS is not equal to PQ then this is not a square, a square is suppose to have equal sides

But I suspect one of the vertices is wrong, vertices S it should have been (-1,-8) and not (-1,-18)

So using S(-1,-8)

Let apply this to the slope

Q(4, 3), S(-1, -8)

m2=∆y/∆x

m2=(y2-y1)/(x2-x1)

m2=(-8-3)/(-1-4)

m2=-11/-5

m2=11/5

Now,

Let find the negative reciprocal of m2

Reciprocal of m2 is 5/11

Then negative of it is -5/11

Which is equal to m1

Then, the square diagonal is perpendicular

6 0

3 years ago