HELP!! I know how to find the answer but I don't know how to use the number line to find the answer.

Answer:

The answer is yes. Since there is no x^2

Answer: c

Step-by-step explanation:

The measure of  angle m∠PZQ is 63degrees.

A point where two or more line segments meet is called a vertex.

The vertex point also forms an angle.

Therefore,

point P is in the interior of ∠OZQ,

Hence,

<OZQ = <OZP +  m∠PZQ

The following angles are given:

m∠OZQ = 125°

m∠OZP = 62°

Substitute the given values into the expression above:

125 = 62 +  m∠PZQ

m∠PZQ = 125 - 62

m∠PZQ = 63°

Therefore, the measure of  the angle m∠PZQ is 63degrees

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Answer:

x =20°

because m and n is parallel

Answer:

Reflected function is y = - x³ + 3

Step-by-step explanation:

When a function f(x) is reflected about the y-axis, the reflected function becomes g(x) = f(-x)

f(x) = x³ + 3

f(-x) = (-x)³ + 3 =-x³ + 3

The amount that would be in the account kept with BPI bank in 3 years is P3,011.2640

It would 6.9422 years for the deposit to double if interest rate is 10,5%

What is future value of immediate amount?

The future value of deposit today in 3 years means its future equivalent when the amount has earned interest over the 3 years period

FV=PV*(1+r)^N

FV=future value after 3 years=unknown

PV=immediate deposit=3,000

r=annual interest rate=0.125%

N=number of years that the deposit lasted=3

FV=3000*(1+0.125%)^3

FV=P3,011.2640

Time taken to double:

FV=future value=3000*2=6000

PV=3000

r=annual interest rate=10.5%

N=number of years that it takes initial deposit to double=unknown

6000=3000*(1+10.5%)^N

6000/3000=(1.105)^N

2=(1.105)^N

take log of both sides

ln(2)=N*ln(1.105)

N=ln(2)/ln(1.105)

N=6.9422 years

Annuity:

FV=PMT*(1+r)^N-1/r

FV=future value of annuity=unknown

PMT=annual payment=20000

r=interest rate=3%

N=number of annual payments=7

FV=20000*(1+3%)^7-1/7%

FV= 153,249.2436

FV=PMT*(1+r)^N-1/r*(1+r)

FV=20000*(1+3%)^7-1/7%*(1+7%)

FV= 157,846.7209

20,000,000=1,250,000*(1+r)^20

20,000,000/1,250,000=(1+r)^20

16=(1+r)^20

(16)^1=(1+r)^20

divide indexes on both sides by 20

(16)^(1/20)=1+r

r=(16)^(1/20)-1

r=14.8698%

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Reason: The absolute value erases the negative sign.

Saying |-29| = 29 means that the number -29 is 29 units from 0 on the number line.

Answer: 29

Step-by-step explanation: The absolute value of -29 is 29. An absolute value is a number's distance from 0. -29 + 29 = 0, therefore the absolute value is 29.

Answer:

y = - 8x + 3

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 8x - 2 ← is in slope- intercept form

with slope m = - 8

• Parallel lines have equal slopes , then

y = - 8x + c ← is the partial equation of the parallel line

to find c substitute (- 0.5, 7 ) into the partial equation

7 = 4 + c ⇒ c = 7 - 4 = 3

y = - 8x + 3 ← equation of parallel line

Answer:

A graphing calculator can help you.

Here is the graph, it is in the photo.

Answer:

- 1

Step-by-step explanation:

2x + 2y = 1

=> 2y = - 2x + 1

=> y = - x + 1

=> Slope = - 1

The length of the field with area 119 square feet is 17 feet and the width of the feet is 7 feet.

The area is the total region or space in two dimension that is covered by a figure , surface or object. Area of a rectangle is calculated by the product of its length and width.

A rectangle has 4 sides where each pair of  opposite side is equal .The length of a rectangle is normally the longer side and the width defines the shorter side.

Given the area of the barn is 119 square feet.

Let us consider the width of the barn to be x feet. As the length is 10 feet longer than the width then we will find the length to be ( x +10 ) feet.

Area = length × width

or, 119 = x × (x+10)

or, 119 = x² +10x

or, x² +10x - 119=0

Now we will solve the quadratic equation thus formed by middle term-factorization method:

or, x² +17x - 7x - 119 = 0

or, x ( x + 17 ) - 7 ( x + 17) = 0

or, ( x + 17 ) ( x - 7 ) = 0

Now either x + 17 = 0 or x - 7 =0

Therefore either x = 7  

or, x = -17 (The width cannot be a negative value)

∴Width = 7 feet and length = 7 + 10 = 17 feet.

Therefore the length of the field is 17 feet and the width of the feet is 7 feet.

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Using Venn sets, the cardinalities are given as follows:

a) n(A U B) = 15.

b) n(A' U C) = 16.

c) n(A ∩ B)' = 10.

Venn amounts relates the cardinality of sets that intersect with each other.

For this problem, the sets are the ones given in this problem, A, B and C, while U is the universal set.

For this problem, the cardinalities are given as follows:

Hence:

Hence:

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Answer:

[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]

Step-by-step explanation:

Given function:

[tex]f(x)=\dfrac{e^x}{\sqrt{e^{2x}+1}}[/tex]

The domain of the given function is unrestricted:  {x : x ∈ R}

The range of the given function is restricted:  {f(x) : 0 < f(x) < 1}

To find the inverse of a function, swap x and y:

[tex]\implies x=\dfrac{e^y}{\sqrt{e^{2y}+1}}[/tex]

Rearrange the equation to make y the subject:

[tex]\implies x\sqrt{e^{2y}+1}=e^y[/tex]

[tex]\implies x^2(e^{2y}+1)=e^{2y}[/tex]

[tex]\implies x^2e^{2y}+x^2=e^{2y}[/tex]

[tex]\implies x^2e^{2y}-e^{2y}=-x^2[/tex]

[tex]\implies e^{2y}(x^2-1)=-x^2[/tex]

[tex]\implies e^{2y}=-\dfrac{x^2}{x^2-1}[/tex]

[tex]\implies \ln e^{2y}= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies 2y \ln e= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies 2y(1)= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies 2y= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

[tex]\implies y= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

Replace y with f⁻¹(x):

[tex]\implies f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]

The domain of the inverse of a function is the same as the range of the original function.  Therefore, the domain of the inverse function is restricted to {x : 0 < x < 1}.

Therefore, the inverse of the given function is:

[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]

The rule f ( x ) - g ( x ) is given as x^3 - x^2 + 18

Given data

f(x) = x3 + 10, g(x) = x2 - 8, and h(x) = 5x + 4

f(x) - g(x) = ( x3 + 10 ) - ( x2 - 8 )

= ( x3 + 10 ) - ( x2 - 8 )

opening the parenthesis gives

= x3 + 10 - x2 + 8

collecting like terms gives

= x3 - x2 + 18

Hence we can say that f(x) - g(x) = x3 - x2 + 18

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complete question

The complete question is attached

(A) (m,n) = g(n,m) for all positive integers m,n,

(B) (m,n + 1) = g(m + 1,n) for all positive integers m,n,

(D) (2m,2n) = (g(m,n))2 for all positive integers m,n

[tex] \rm f(m,n,p) = \sum \limits_{i = 0}^{p} {}^{m} C_{i} \: \: {}^{n + i} C_{p} \: \: {}^{p + n} C_{p - i}[/tex]

[tex]\rm {}^{m} C_{i} \: \: {}^{n + i} C_{p} \: \: {}^{p + n} C_{p - i}[/tex]

[tex]\rm {}^{m} C_{i} \dfrac{(n + i)!}{p!(n - p + i)!} \times \dfrac{(n + p)!}{(p - i)!(n + i)!} [/tex]

[tex]\rm {}^{m} C_{i} \times \dfrac{(n + p)!}{p!} \times \dfrac{1!}{(n -p + i)!(p - i)!} [/tex]

[tex]\rm {}^{m} C_{i} \times \dfrac{(n + p)!}{p!} \times \dfrac{1!}{(n -p + i)!(p - i)!} [/tex]

[tex]\rm {}^{m} C_{i} \times \dfrac{(n + p)!}{p!n!} \times \dfrac{n!}{(n -p + i)!(p - i)!} [/tex]

[tex]\rm {}^{m} C_{i} \: \: {}^{n + p} C_{p} \: \: {}^{n} C_{p - i} \: \: \{{}^{m} C_{i} \: \: {}^{n } C_{p - i} = {}^{m + n} C_{p } \}[/tex]

[tex]\rm {}^{m} C_{i} \: \: {}^{n + p} C_{p} \: \: {}^{n} C_{p - i} \: \: \{{}^{m} C_{i} \: \: {}^{n } C_{p - i} = {}^{m + n} C_{p } \}[/tex]

[tex] \rm f(m,n,p) = {}^{n + p} C_{p}{}^{m + n} C_{p}[/tex]

[tex] \rm \dfrac{f(m,n,p)}{{}^{n + p} C_{p}} = {}^{m + n} C_{p}[/tex]

Now,

[tex] \rm g(m,n) = \sum \limits_{p = 0}^{m + n} \dfrac{f(m,n,p)}{{}^{n + p} C_{p}} [/tex]

[tex] \rm g(m,n) = \sum \limits_{p = 0}^{m + n} {}^{m + n} C_{p}[/tex]

[tex] \rm g (m,n) = {2}^{m + n} [/tex]

[tex] \rm(A) \: g(m,n) = q(n,m)[/tex]

[tex] \rm(B) \: g(m,n + 1) = {2}^{m + n + 1} [/tex]

[tex] \rm g(m + n,n ) = {2}^{m + 1 + n} [/tex]

[tex] \rm(D) \: g(2m,2n) = {2}^{2m + 2n} [/tex]

[tex] = \rm( {2}^{m + n} {)}^{2} [/tex]

[tex] = \rm(g(m,n)) {}^{2} [/tex]

The number of teachers that work in both schools is 97.

The addition is a way of combining things and counting them together as one large group.

We have,

Number of teachers in Alberts Elementary school = 76

Number of teachers in Bothel Elementary school = 24

Number of teachers that work in both schools = 3

The number of teachers that work in both schools:

= 76 + 24 - 3

We are subtracting 3 from the total because 3 teachers that work in both schools have been added twice.

= 100 - 3

= 97

Thus the number of teachers that work in both schools is 97.

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infinitely many solutions 

Example

Simplify left side using distributive property to 6x+16. Combine like terms (2x and 4x) on the right side to 6x+16. Both sides simplify to the same expression (left side = right side) so there are infinitely many solutions. You can plug in any real number for x and the left side will always equal the right side.

Answer: Yes

Step-by-step explanation: Kelly's results are likely to be correct mainly because she pulled votes from several different age groups. The fact that she asked several age groups whether or not they would be in favor of creating a local dog park shows that her results are likely to yield valid results.

Answer:

[tex]\textsf{1)} \quad t=\dfrac{48}{r}[/tex]

[tex]\textsf{2)} \quad \sf 2.4\:hours=2\:hours\:24\:minutes[/tex]

Step-by-step explanation:

[tex]\boxed{\sf Time=\dfrac{Distance}{Speed}}[/tex]

Given:

Substitute the given values into the formula:

[tex]\implies t=\dfrac{48}{r}[/tex]

Substitute r = 20 km/h into the derived equation from part 1 and solve for t:

[tex]\implies t=\dfrac{48}{20}=2.4\: \sf hours[/tex]

Note: 2.4 hours = 2 hours 24 minutes

Answer: 2.4 hours

Step-by-step explanation: t=time

         r=rate        so question one is equal to t=48/r

then question 2 is 48/20=2.4

Answer:-

Paul has $64 pocket money each day.

Explanantion:-

Let amy ,flora and Paul 's pocket money be x,y and z respectively.

So,

By the question :

x = 3y

So, y = x/3 - eqn i

z = 7 + x - eqn ii

7(x + y + z) = 980

x + y + z = 140

Now,

x + y+ z = 140

x + x/3 + 7 + x = 140

or, 2x + x/3 = 133

or, 6x + x = 399

so, x = 399/7

so, x = $57

Now,

z = 7 + x

= 7 + $57

= $64

Answer:

X= -7

X = 3

Step-by-step explanation:

| x + 2| = 5

|-7 + 2| = 5

|-5| = 5

5 = 5

|x + 2| = 5

|3 + 2| = 5

|5| = 5

5 = 5

X= -7

X = 3

Yes; for each element of the domain, there is only one element of the range.

For a relation in coordinate form to be a function, there is must nor be repetition in the value of its coordinate.

Given the following relation as shown  { ( 2 , 5 ) , ( 4 , − 2 ) , ( 3 , 3 ) , ( 5 , 4 ) , ( − 2 , 5 ) }, since there is no repetition in its domain values, hence the relation is a function.

We can conclude therefore that Yes; for each element of the domain, there is only one element of the range.

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Step-by-step explanation:

wot. that's not even a qn bro

Step-by-step explanation:

the difference between the numbers are 5

so this is an arthimetic sequence with D=5

5n-7 is the formula

Answer:

x ≥ 3

Step-by-step explanation:

2(x - 3) ≥ -3(-3 + x)

2x - 6 ≥ 9 - 3x

5x - 6 ≥ 9

5x ≥ 15

x ≥ 3

Answer:

x ≥ 3

Step-by-step explanation:

write, calculate, divide both sides

[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

Domain : All possible values of x for which a function is defined.

According to given graph, the curve extends to infinity on both sides on x - axis, so it is defined for all real values.

i.e : Domain : All real

Range : All possible values of y for x, as per the given function

As per the given graph, the graph extends to infinity on lower end, but has maximum value of 0.

i.e Range : [tex]\boxed{ \sf y < 0} [/tex]

Answer:

(x/8)(19 - x)

Step-by-step explanation:

The quotient is the answer to a division problem.

The difference is the answer to a subtraction problem.

"Times" indicates multiplication, which you could look at it as a link between the two functions, or quantities (them being the constants and variables sectioned off into parenthesis).