when simplifying the expression y=(2x(x-3)(x-3))/(x-1)(x-3) do all of the x-3 s get cancelled or just one in the numerator and one in the denominator?​

X - (-20) = 5

When you subtract a negative, change it to addition:

X + 20 = 5

Subtract 20 from both sides:

X = -15

Answer:

[tex]\boxed{x=-15}[/tex]

Step-by-step explanation:

[tex]x-(-20)=5[/tex]

[tex]\sf Distribute \ negative \ sign.[/tex]

[tex]x+20=5[/tex]

[tex]\sf Subtract \ 20 \ from \ both \ sides.[/tex]

[tex]x+20-20=5-20[/tex]

[tex]x=-15[/tex]

Answer:

E) 15pi/2

Step-by-step explanation:

The total circle area is pi*r^2. r = 6. so the total area is 6^2pi, or 36pi.

However, we are dealing with 75/360 of a circle, or 5/24 of a circle. 5/24 * 36pi = 15pi/2.

Answer:

Option (1)

Step-by-step explanation:

Given equation is,

[tex]2^x=1-3^x[/tex]

To determine the solution of the equation we will substitute the values of 'x' given in the options,

Option (1)

For x = -0.75

[tex]2^{-0.75}=1-3^{-0.75}[/tex]

0.59 = 1 - 0.44

0.59 = 0.56

Since, values on both the sides are approximately same.

Therefore, x = -0.75 will be the answer.

Option (2)

For x = -1.25

[tex]2^{-1.25}=1-3^{-1.25}[/tex]

0.42 = 1 - 0.25

0.42 = 0.75

Which is not true.

Therefore, x = -1.25 is not the answer.

Option (3)

For x = 0.75

[tex]2^{0.75}=1-3^{0.75}[/tex]

1.68 = 1 - 2.28

1.68 = -1.28

Which is not true.

Therefore, x = 0.75 is not the answer.

Option (4)

For x = 1.25

[tex]2^{1.25}=1-3^{1.25}[/tex]

2.38 = 1 - 3.95

2.38 = -2.95

It's not true.

Therefore, x = 1.25 is not the answer.

Answer:

B. f(n) = 56(0.5)^n-1

Step-by-step explanation:

 First, You have to find out the starting population, if you look at the problem you see the population starts at 56

f(x) = 56

Second, you know that the population goes down 50% each week so it has a decay of 0.5

f(x) = 56(0.5)

 Third,  you need to add the exponent of n to make it exponential.  But, if you just add n then the the population would be 28 on week 1 which is incorrect. To fix that you make the exponent n-1 so when you are on week 1 it doesn't become 28 but it stays on 56, and on week 2 it's 28, ect

f(x) = 56(0.5)^n-1

Answer:

143.32 m

Step-by-step explanation:

Given the following :

Diameter of circular track = 80m

Hillary's speed = 300m per minute

Eugene's speed = 240m per minute

Run time = 50 minutes

Note: they both run in opposite direction.

Calculate the Circumference(C) of the circle :

C = 2πr or πd

Where r = radius ; d = diameter

Using C = πd

C = πd

C = π * 80

C = 251.327

Eugene's distance covered = (240 * 50) = 12000

Hillary's distance covered = (300 * 50) = 15000

Number turns :

Eugene = 12000/ 251.327 = 47.746561

Hillary = 15000/251.327 = 59.683201

Therefore ;

48 - 47.746561 = 0.253439

60 - 59.683201 = 0.316799

(0.253439+0.316799) = 0.570238

Distance which separates Eugene and Hillary when they finish :

0.570238 * 251.327 = 143.32 m

Answer:

2¹ + 3¹, 2³ - 2¹, 3²

Step-by-step explanation:

2³ - 2¹ = 8 - 2 = 6

2¹ + 3¹ = 2 + 3 = 5

3² = 9

Answer:

Step-by-step explanation:

The triangle inequality theorem states that the sum of any two sides of a triangle os greater than the third side.

■■■■■■■■■■■■■■■■■■■■■■■■■■

First triangle:

Let a,b and c be the sides of the triangle:

● a = 10

● b = 20

● c = 30

Now let's apply the theorem.

● a+b = 10+20=30

That's equal to the third side (c=30)

●b+c = 50

That's greater than a.

● a+c = 40

That's greater than b.

These aren't the sides of a triangel since the first inequality isn't verified.

■■■■■■■■■■■■■■■■■■■■■■■■■

Second triangle:

● a = 122

● b = 257

● c = 137

Let's apply the theorem.

● a+b = 379

That's greater than c

● a+c = 259

That's greater than b

● b+c = 394

That's greater than a

So 122,257 and 137 can be sides of a triangle.

■■■■■■■■■■■■■■■■■■■■■■■■■■

The third triangle:

● a = 8.6

● b = 12.2

● c = 2.7

Let's apply the theorem:

● a+b = 20.8

That's greater than c

● b+c = 14.9

That's greater than a

● a+c = 11.3

That isn't greater than b

So theses sides aren't the sides of triangle.

■■■■■■■■■■■■■■■■■■■■■■■■■■

● a = 1/2

● b = 1/5

● c = 1/6

Let's apply the theorem.

● a+b = 7/10

That's greater than c

● a+c = 2/3

That's greater than b

● b+c = 11/30

That isn't greater than a

So these can't be the sides of a triangle.

Answer:

d. 3√6 = 7.348

Step-by-step explanation:

1. simplify each expression

a. √150 / 2 = 6.124

b. π + 4 = 7.142

c. 2π = 6.283

d. 3√6 = 7.348

the largest number will be the closest to 8. therefore, point W is expression D.

Answer:

Step 2

Step-by-step explanation:

9 was added to both sided so the equation would remain equal and the 9 would be cancelled out on the left side.

Answer:

B. -75

Step-by-step explanation:

you count backwards from 5, 27 times

Answer:

809.915$

Step-by-step explanation:

Amount of money = Principal x e^(rate x year)

                              = 600 x e^(0.05 x 6)

                              = 809.915$

Answer:

$809.92

Step-by-step explanation:

(see attached for reference)

Recall that the formula for compound interest (compounded continuously) is

A = P e^(rt)

where,

A = final amount (we are asked to find this)

P = principal = given as $600

r = interest rate = 5% = 0.05

t = time = 6 years

e = 2.71828 (mathematical constant)

Substituting the known values into the equation:

A = P e^(rt)

= 600 e^(0.05 x 6)

= 600 (2.71828)^(0.30)

= $809.92

Answer:

(3, 2) and (1, 4)

Step-by-step explanation:

Plot the two points on a graph.

The other two points are (3, 2) and (1, 4).

To do this with algebra, it takes a few steps.

The diagonals of a square are perpendicular and bisect each other. You are given opposite vertices, so first, find the midpoint of that diagonal.

((1 + 3)/2, (2 + 4)/2) = (2, 3)

The midpoint of the diagonal is (2, 3).

This diagonal has slope 1 and y-intercept 1, so its equation is

y = x + 1

The perpendicular bisector has equation

y = -x + 5

The two vertices we are looking for, lie in a circle whose center is the midpoint of the diagonals, (2, 3), and whose radius is half of the diagonal.

Use Pythagoras to find the diagonal's length.

2^2 + 2^2 = c^2

c^2 = 8

c = sqrt(8) = 2sqrt(2)

Half of the diagonal is sqrt(2). This is the radius if the circle.

The equation of the circle is

(x - 2)^2 + (y - 3)^2 = (sqrt(2))^2

(x - 2)^2 + (y - 3)^2 = 2

The points of intersection of this circle and the second diagonal are the two vertices you are looking for.

System of equations:

(x - 2)^2 + (y - 3)^2 = 2

y = -x + 5

Use substitution and substitute y with -x + 5 in the equation of the circle.

(x - 2)^2 + (-x + 5 - 3)^2 = 2

(x - 2)^2 + (-x + 2)^2 - 2 = 0

x^2 - 4x + 4 + x^2 - 4x + 4 - 2= 0

2x^2 - 8x + 6 = 0

x^2 - 4x + 3 = 0

(x - 3)(x - 1) = 0

x - 3 = 0   or x - 1 = 0

x = 3 or x = 1

Now we find corresponding y values.

y = -x + 5

x = 3

y = -3 + 5 = 2

This gives us (3, 2).

y = -x + 5

x = 1

y = -1 + 5 = 4

This gives us (1, 4).

Answer: (1, 4) and (3, 2)

AC will be 4√2 when AB = 4 and BC = 4, in the given right triangle.

According to Pythagoras' Theorem, in a right triangle, the square of the length of the longest side, that is, the hypotenuse, that is, the side opposite to the right angle is equal to the sum of the squares of the lengths of the other two sides.

In the question, we are given a right triangle, with sides AB = 4 and BC = 4.

We are asked to find AC.

To find AC, we will use the Pythagoras theorem, according to which, we can write:

AC² = AB² + BC²

or, AC² = 4² + 4²,

or, AC² = 16 + 16,

or, AC² = 32,

or, AC = √32,

or, AC = √(16 * 2) = 4√2.

Therefore, AC will be 4√2 when AB = 4 and BC = 4, in the given right triangle.

Learn more about Pythagoras' Theorem at

https://brainly.com/question/231802

#SPJ2

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

2(6y - 5) = 2

12y - 10 = 2

12y = 12

y = 1

x = 6(1) - 5

x = 6 - 5 = 1

(1,1)

Answer: (1,1)

Step-by-step explanation: 2(6y - 5) = 2

12y - 10 = 2

12y = 12

y = 1

x = 6(1) - 5

x = 6 - 5 = 1

(1,1)

Answer:

x = 19 1/3

Step-by-step explanation:

6/14 = 7/(x-3)

Using cross products

6 * (x-3) = 7*14

6x -18 = 98

Add 18 to each side

6x-18+18 =98+18

6x = 116

Divide each side by 6

6x/6 = 116/6

x =58/3

x = 19  1/3

Answer= 58/3

Step by Step

Step 1: Cross-multiply.

6*(x−3)=(7)*(14)

6x−18=98

Step 2: Add 18 to both sides.

6x−18+18=98+18

6x=116

Step 3: Divide both sides by 6.

6x /6 = 116 /6

x= 58 /3

Answer:

Domain is all real numbers, x ≠ -5

Range is all real numbers, y ≠ 0

Step-by-step explanation:

Answer:

x = 9/4

y = 3/5

z = 2/3

w = -9/5

Step-by-step explanation:

Technically, the matrix is in reduced row echelon form. If there are zeros above and below the ones, it is RREF. If there are zeros only below the ones, then it's REF.

Since it is in RREF, the augmented numbers to the right of the bar are already your solutions. Simply label the variables.

Answer:

F , D, E

Step-by-step explanation:

The smallest angle is opposite the smallest side

The largest angle is opposite the largest side

DE is smallest so F is smallest

Then EF so D

DF is largest so E is the largest angle

Answer:

The steps to construct a a line parallel to another line from a point includes

1) From the given line draw a transversal through the point

2) With the compass, copy the angle formed between the transversal and the given line to the point P

3) Draw a line through the intersection of the arcs of the angle construction to get the parallel line through the point P

Step-by-step explanation:

Answer:

D. [tex]\sqrt{\frac{2*2*2*2*3}{5*5*7} }[/tex]

Step-by-step explanation:

48 divided by 2 is 24. Insert the 2.

24 divided by 2 is 12. Insert the 2.

12 divided by 2 is 6. Insert the 2.

6 divided by 2 is 3. Insert the 2 and 3:

[tex]48=2*2*2*2*3[/tex]

175 divided by 5 is 35. Insert the 5.

35 divided by 5 is 7. Insert the 5 and 7:

[tex]175=5*5*7[/tex]

:Done

Answer:

Proof in the explanation.

Step-by-step explanation:

I expanded both sides as a first step. (You may use foil here if you wish and if you use that term.)

This means we want to show the following:

[tex]3-9\tan^2(\theta)-4\sin^2(\theta)+12\sin^2(\theta)\tan^2(\theta)[/tex]

[tex]=12\cos^2(\theta)-9-4\cos^2(\theta)\tan^2(\theta)+3\tan^2(\theta)[/tex].

After this I played with only the left hand side to get it to match the right hand side.

One of the first things I notice we had sine squared's on left side and no sine squared's on the other. I wanted this out. I see there were cosine squared's on the right. Thus, I began with Pythagorean Theorem here.

[tex]3-9\tan^2(\theta)-4\sin^2(\theta)+12\sin^2(\theta)\tan^2(\theta)[/tex]

[tex]3-9\tan^2(\theta)-4(1-\cos^2(\theta))+12\sin^2(\theta)\tan^2(\theta)[tex]

Distribute:

[tex]3-9\tan^2(\theta)-4+4\cos^2(\theta)+12\sin^2(\theta)\tan^2(\theta)[/tex]

Combine like terms and reorder left side to organize it based on right side:

[tex]4\cos^2(\theta)-1+12\sin^2(\theta)\tan^2(\theta)-9\tan^2(\theta)[/tex]

After doing this, I since that on the left we had products of sine squared and tangent squared but on the right we had products of cosine squared and tangent squared. This problem could easily be fixed with Pythagorean Theorem again.

[tex]4\cos^2(\theta)-1+12\sin^2(\theta)\tan^2(\theta)-9\tan^2(\theta)[/tex]

[tex]4\cos^2(\theta)-1+12(1-\cos^2(\theta))\tan^2(\theta)-9\tan^2(\theta)[/tex]

Distribute:

[tex]4\cos^2(\theta)-1+12\tan^2(\theta)-12\cos^2(\theta)\tan^2(\theta)-9\tan^2(\theta)[/tex]

Combined like terms while keeping the same organization as the right:

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-12\cos^2(\theta)\tan^2(\theta)-9\tan^2(\theta)[/tex]

We do not have the same amount of the mentioned products in the previous step on both sides. So I rewrote this term as a sum. I did this as follows:

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-12\cos^2(\theta)\tan^2(\theta)-9\tan^2(\theta)[/tex]

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-4\cos^2(\theta)\tan^2(\theta)-8\cos^2(\theta)\tan^2(\theta)-9\tan^2(\theta)[/tex]

Here, I decide to use the following identity [tex]\cos\theta)\tan(\theta)=\sin(\theta)[/tex]. The reason for this is because I certainly didn't need those extra products of cosine squared and tangent squared as I didn't have them on the right side.

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-4\cos^2(\theta)\tan^2(\theta)-8\cos^2(\theta)\tan^2(\theta)-9\tan^2(\theta)[/tex]

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-4\cos^2(\theta)\tan^2(\theta)-8\sin^2(\theta)-9\tan^2(\theta)[/tex]

We are again back at having sine squared's on this side and only cosine squared's on the other. We will use Pythagorean Theorem again here.

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-4\cos^2(\theta)\tan^2(\theta)-8\sin^2(\theta)-9\tan^2(\theta)[/tex]

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-4\cos^2(\theta)\tan^2(\theta)-8(1-\cos^2(\theta))-9\tan^2(\theta)[/tex]

Distribute:

[tex]4\cos^2(\theta)-1+3\tan^2(\theta)-4\cos^2(\theta)\tan^2(\theta)-8+8\cos^2(\theta)-9\tan^2(\theta)[/tex]

Combine like terms:

[tex]12\cos^2(\theta)-9+3tan^2(\theta)-4\cos^2(\theta)\tan^2(\theta)[/tex]

Reorder again to fit right side:

[tex]12\cos^2(\theta)-9+4\cos^2(\theta)\tan(\theta)+3\tan^2(\theta)[/tex]

This does match the other side.

The proof is done.

Note: Reordering was done by commutative property.

Step-by-step explanation:

log 10 = 1.  So if log x < 1, then x < 10.  And if log x > 1, then x > 10.

The upper left number is the smallest, and can't be smaller than 1.  If the exponent is 0, we can put any number in the red box.

The fractions in the upper right and lower left need to be as large as possible.  The denominators will be small, and the numerators will be large.

From there, a little trial and error does the rest.  The are many possible answers.  I've included one.

Answer:

I hope it helps.

The solutions in photo ^_^

Answer:

B) Associative Property of Multiplication

Step-by-step explanation:

*if it's wrong idk how, but I apologise*

Answer:

x=27

Step-by-step explanation:

expanding the above expression we get

5x+5=4x+32

grouping numbers with coefficient of x at the left side and constant at the right side we get

5x-4x=32-5

x=27

Answer: 880 cm

Step-by-step explanation:

Given: Radius of the hula hoop = 35 cm

Hula hoop is  circular in shape

Then, Circumference = [tex]2\pi r[/tex] , where r = radius

Now , Circumference of hula hoop = [tex]2\times \dfrac{22}{7}\times35=220\ cm[/tex]

Now , the distance the hula hoop rolls in 4 full rotations = 4 × (Circumference of hula hoop)

[tex]= 4 \times 220=880\ cm[/tex]

Hence, the required distance = 880 cm

Answer:

880

Step-by-step explanation:

===========================================

Explanation:

I'll use x in place of n

Let y = x^2 - 4x + 5

If we complete the square, then,

y = x^2 - 4x + 5

y = (x^2 - 4x) + 5

y = (x^2 - 4x + 4 - 4) + 5

y = (x^2 - 4x + 4) - 4 + 5

y = (x-2)^2 + 1

The quantity (x-2)^2 is never negative as squaring any real number value is never a negative result. Adding on 1 makes the result positive. So y > 0 regardless of whatever x is. Replace x with n, and this shows how n^2 - 4n + 5 is always positive for any integer n.

------------

You could also use the quadratic formula to find that x^2 - 4x + 5 = 0 has no real solutions, so there are no x intercepts. Either the graph is entirely above the x axis or it is entirely below the x axis.

Plug in any x value you want, say x = 0, and the result is positive. Meaning that whatever x value you plug in will be positive (as the graph can't cross the x axis to go into negative territory)

Answer:

10 cm is the answer because 30÷3 angles

Answer:

x = -4

Step-by-step explanation:

logs (8 - 3x) = log20

Since we are taking the log on each side

log a = log b  then a = b

8 -3x = 20

Subtract 8  from each side

8 -3x-8 =20 -8

-3x = 12

Divide by -3

-3x/-3 = 12/-3

x = -4

Answer:

[tex] \boxed{\sf x = -4} [/tex]

Step-by-step explanation:

[tex] \sf Solve \: for \: x \: over \: the \: r eal \: numbers:[/tex]

[tex] \sf \implies log(8 - 3x) = log 20[/tex]

[tex] \sf Cancel \: logarithms \: by \: taking \: exp \: of \: both \: sides:[/tex]

[tex] \sf \implies 8 - 3x = 20[/tex]

[tex] \sf Subtract \: 8 \: from \: both \: sides:[/tex]

[tex] \sf \implies 8 - 3x - 8 = 20 - 8 [/tex]

[tex] \sf \implies - 3x = 12 [/tex]

[tex] \sf Divide \: both \: sides \: by \: - 3:[/tex]

[tex] \sf \implies \frac{-3x}{-3} = \frac{12}{-3} [/tex]

[tex] \sf \implies x = - 4[/tex]