For the graph y = 1 find the slope of a line that is perpendicular to it and the slope of a line parallel to it. Explain your answer with two or more sentences.

Answer:

1. a)

[tex] \begin{gathered} \frac{2}{ \sqrt{3} - 1 } = \frac{2}{ \sqrt{3} - 1} \times \ \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } \\ = \frac{2 \sqrt{3} + 2 } {( \sqrt{3} ) {}^{2} - 1 {}^{2} } = \frac{2 \sqrt{} 3}{3 - 1} = \frac{2 \sqrt{3} }{2} \\ = \sqrt{3} \end{gathered}[/tex]

[tex]\\[/tex]

b)

[tex]\begin{gathered} = > \: \: \frac{7}{ \sqrt{12} - \sqrt{5} } \\ \\ = > \: \: \frac{7}{ \sqrt{12} - \sqrt{5} } \times \frac{ \sqrt{12} + \sqrt{5} }{ \sqrt{12} + \sqrt{5} } \\ \\ = > \: \: \frac{7( \sqrt{12} + \sqrt{5} ) }{ {( \sqrt{12}) }^{2} - {( \sqrt{5}) }^{2} } \\ \\ = > \: \: \frac{7( \sqrt{12} + \sqrt{5}) }{12 - 5} \\ \\ => \: \: \frac{ \cancel{7}( \sqrt{12} + \sqrt{5} ) }{ \cancel{7}} \\ \\ => \: \: \sqrt{12} + \sqrt{5} \end{gathered}[/tex]

1-:

[tex] \displaystyle \frac{2}{ \sqrt{3} - 1 } [/tex]

[tex] \displaystyle = \frac{2}{( \sqrt{3} - 1)( \sqrt{3} + 1 } \\ [/tex]

[tex] \displaystyle2 \frac{ \sqrt{3} + 1 }{ \sqrt{3 {}^{2} - 1 {}^{2} } } [/tex]

[tex] \displaystyle2 \frac{ \sqrt{3} + 1}{3 - 1} [/tex]

[tex] \displaystyle \: \sqrt{3} + 1[/tex]

[tex] \displaystyle2 - ) \frac{7}{ \sqrt{12} - \sqrt{5} } [/tex]

[tex] \displaystyle\frac{ \sqrt{12} \sqrt{5} }{ \sqrt{12 {}^{2} } \times \sqrt{5 {}^{2} } } [/tex]

[tex] \displaystyle7\frac{ \sqrt{12} + \sqrt{5} }{7} = \sqrt{12} + \sqrt{5} [/tex]

[tex] \displaystyle3) = \frac{8 + 3 \sqrt{5} }{64 - 45} \\ = \frac{8 - 3 \sqrt{5} }{19} [/tex]

Answer:

2

Step-by-step explanation:

The graphs intersect at 2 points.

The 2 solutions are x = -2 and x = 6.

The value of f from the given expression is 3

Given the expression below

cos(22f-1) = sin(7f+4),

This can also be expressed

cos(22f-1) = cos(90-7f-4)

cos(22f-1) = cos(86-7f)

22f - 1 = 86 - 7f

Collect the like terms

22f+7f = 86 + 1

29f = 87

f = 3

Hence the value of f from the given expression is 3

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If CD is perpendicular on AB then the x intercept of CD is (17,0).

Given that CD is perpendicular on AB and passes through C(5,12) and the coordinates of A and B are  (-10,-3), (7,14).

When two lines are perpendicular on other line then the product of their slopes is equal to -1.

It is given that CD is perpendicular on AB. So we need to first calculate the slope of AB.

Slope=[tex](y_{2} -y_{1} /x_{2} -x_{1} )[/tex]

Slope of AB=(14+3/7+10)

=(17/17)

=1

According to rule product of slopes of AB and CD should be -1.

So the slope of CD will be -1.

Now we have to form equation of CD from slope -1 and point (5,12).

y-12=-1(x-5)

y-12=-x+5

x+y-17=0

We have to find out x intercept.

We know that x intercept is a point where the line touches x axis. So here the value of y =0.

x+y-17=0

put the value of y=0.

x+0-17=0

x=17

Point will be (17,0)

Hence the x intercept of CD will be (17,0).

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Question is incomplete as it should include in its beginning that CD is perpendicular on AB.

The value of test statistic is Z =  -2.42.

According to the statement

we have given that the person in the US uses 123 gallons of water per day.

And conducted a survey of 40 randomly selected Americans and get a mean value is 113.03 gallons and a standard deviation of 25.99 gallons.

And we have to test the hypotheses with test statistic.

So,

Here we use the the z- test for one population =

Z = (sample mean - Claimed hypothesis mean)/ (standard deviation/(sample size)^1/2.)

Here the sample mean = 113.03

And claimed hypothesis mean = 123

Standard deviation = 25.99

Sample size = 40

Now put the values in the above written formula is

Z = (sample mean - Claimed hypothesis mean)/ (standard deviation/(sample size)^1/2.)

Z = 113.03 - 123 / (25.99/(40)^1/2)

Z = -9.97 / (25.99 / 6.32)

Z = -9.97 / 4.11

Z =  -2.42

So, The value of test statistic is Z =  -2.42.

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If the value of μ is 65, σ is 5 and z is 1.14 then the value of possible voter age would be 70.6 years.

Given that value of μ is 65, σ is 5 and z is 1.14.

We have to calculate the possible voter age that can satisfy the given values of σ,μ and z.

We know that in z test,

z=X-μ/σ

in which μ is population mean,σ is population standard deviation.

We have to just put the values in the above formula to get the value of X which is our possible voter age.

Putting the values we get,

1.14=x-65/5

5.7=X-65

X=65+5.7

X=67.7

Hence the possible voter age is 67.7 which would satisfy the value of μ=65,σ=5 and z=1.14.

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The graph is a decreasing exponential graph.

Part A: The given graph is decreasing the exponential graph since the graph is decreasing rapidly as the value of x is increasing and the value of y is decreasing rapidly.

Part B: The graph shown here is maybe the case of any electric appliance which is of cheap quality. The time when it is brought then its performance and its life is awesome after that its life is decreasing rapidly.

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Complete question:

Describe a relationship that can be modeled by the function represented by the graph, and explain how the function models the relationship. Identify and interpret the key features of the function in the context of the situation you described in part A.

To solve a triangle means to obtain al the missing parts of the triangle.

To solve a triangle means to obtain al the missing parts of the triangle. Now we have the following information;

<ABC = 51°

c = 8

a = 11

Then;

b^2 = a^2 + c^2 - 2acCos B

b^2 = (11)^2 + (8)^2 - [2 * 11 * 8 * cos 51]

b^2 = 121 + 64 - 55.38

b = 11.39

Now;

a/sinA = b/sinB

asinB = bsinA

sinA = asinB/b

sinA = 11 * sin 51/11.39

A = 48.6 or 49°

C = 180 - (49 + 51)

C = 80°

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After 39 years, the value of this quantity is 1,023.35

Here we know that:

Then the exponential equation is:

[tex]A(t) = 570*(1 + 0.35)^{t/20}[/tex]

Where t is the time in years.

So, after 39 years, the value of this quantity is given by:

[tex]A(39) = 570*(1 + 0.35)^{39/20} = 1,023.35[/tex]

After 39 years, the value of this quantity is 1,023.35

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The  sum of the links of the two parallel sides is 5cm

The formula for calculating the area of a trapezoid is expressed as:

A = 0.5(a+b)h

where

h is the height of the trapezoid

a+b is the sum of the sides

Substitute

20 = 0.5(a+b)8

40 = 8(a+b)

a+b = 40/8

a+b = 5cm

Hence the  sum of the links of the two parallel sides is 5cm

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The commission rate exists at $8,493 with a rate of 9.5%.

Jorge earns 9% from all his sales, so his commission exists

89400 × 0.09 = $8046

Harris made $447 more than that, so his commission exists

8046 + 447 = $8493

To estimate his commission rate, we can utilize the subsequent equation

89400 × rate percentage = 8493

Rate percentage = 8493 / 89400

Rate percentage = 0.095

Rate percentage = 9.5%

So Harris' commission exists at $8,493 with a rate of 9.5%.

Therefore, the commission rate exists at $8,493 with a rate of 9.5%.

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table 1, and table 2 are the answers.

we know that any function is increasing when for an increase in x-values, y-value should also increase

we will verify each table

table-1:

we can see that

x increase as -3, -2, -1 , 0, 1,2

y is also increasing -6,-3,-1,1,3,6

so, this is increasing

table-2:

we can see that

x increase as -3, -2, -1 , 0, 1,2

but it is asking between -2 to 1

y-values are -4,-1,1,4.....which is increasing

so, this is  increasing

table-3:

we can see that

x increase as -3, -2, -1 , 0, 1,2

but y changes from -3 to -5 .....which is decreasing

so, this is not increasing

table-4:

we can see that

x increase as -3, -2, -1 , 0, 1,2

but y changes from 7 to 1 .....which is decreasing

so, this is not increasing

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Using the combination formula, the probability that a person with one combination of five numbers will win is:

[tex]\frac{1}{1,221,759}[/tex]

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

For this problem, 5 numbers are taken from a set of 45, hence the number of combinations is:

[tex]C_{45,5} = \frac{45!}{5!40!} = 1,221,759[/tex]

Hence the probability is:

[tex]\frac{1}{1,221,759}[/tex]

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Answer:

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

Yep, what you did there is correct!

The expression was: [tex](\frac{5^2}{4} )^4[/tex]

So let's solve this step by step.

First the numerator

[tex](5^2)^4[/tex]

Is the same thing as

[tex]5^8[/tex]

Since the exponents are multiplied.

Next is the denominator
[tex]4^4[/tex]

This can stay the same, since solving it as 4 x 4 x 4 x 4 will only make the fraction more complicated. Also, the answer choices below show that the denominator stays in exponential form.

==> [tex]\frac{5^8}{4^4}[/tex] is our final answer!

So the answer to this question is D. You got the question correct! :)

I hope that helped!!

The complete sentence for the unit circle is:

When t = 0, AP = 0, so sin t will begin at a value of 0. As t increases from 0 to π/2, sin t increases. Eventually, when t = π/2, OP will be 0, so sin t = AP will be 1.

In this problem we have sentence to be completed by direct observation to the variables seen in a unit circle, a form used to teach trigonometric functions in a graphic manner. Now we proceed to present the complete sentence below:

When t = 0, AP = 0, so sin t will begin at a value of 0. As t increases from 0 to π/2, sin t increases. Eventually, when t = π/2, OP will be 0, so sin t = AP will be 1.

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area = pi * r^2

here, the radius is 2 as its the diameter/2.

so the area is 4pi.

circumference = 2*pi*r.

so the circumference is 4pi as well.

Answer:round to nearest hundredth (4,786.3)

Step-by-step explanation:

hope this helps, if wrong im so sorry :(

A kite is a quadrilateral with four sides that may be divided into two adjacent pairs of equal-length sides. The total length of ribbon that is needed to line the two long sides of the hat is 22 inches.

According to the question,

A kite is a quadrilateral with four sides that may be divided into two adjacent pairs of equal-length sides. A parallelogram, on the other hand, has two sets of equal-length sides that are opposite each other rather than adjacent.

Since the two adjacent small and two adjacent long sides of a kite are always equal. Therefore, the length of the smaller sides can be written as,

(2x + 1) inches = 5 inches

2x + 1 = 5

2x = 4

x = 2

Now, the length of the longer side of the kite can be written as,

(3x + 5) inches

= 3(2) + 5

= 11 inches

As both the long sides are needed to be covered with the ribbon, therefore, the length of the ribbon needed is,

Ribbon needed = 11 inches × 2

                         = 22 inches

Hence, a kite is a quadrilateral with four sides that may be divided into two adjacent pairs of equal-length sides. the total length of ribbon that is needed to line the two long sides of the hat is 22 inches.

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The expansion of  (x – 3)^5 is     [tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

The Binomial Theorem is the method of expanding an expression that has been raised to any finite power. A binomial Theorem is a powerful tool of expansion, which has application in Algebra, probability, etc. Binomial Expression: A binomial expression is an algebraic expression that contains two dissimilar terms.

Pascal's Triangle is a never-ending equilateral triangle in which the arrays of numbers arranged in a triangular manner. The triangle starts at 1 and continues placing the number below it in a triangular pattern

Using Binomial theorem,

=[tex]\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}\left(-3\right)^i[/tex]

=[tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

Using Pascal Triangle  for (x-1)

                                               1

                                       1    1              

           1    2    1            

        1    3    3    1        

     1    4    6    4    1      

  1    5    10    10    5    1  

1    6    15    20    15    6    1

Accordingly replacing 1 with 3 we get

= [tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

Thus the expansion of  (x – 3)^5 is      [tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

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Answer:

3. Completing the Square

4. 0, 1, or 2

Step-by-step explanation:

Hello!

a) Graphing

Graphing is a useful method as it lays the answer out in front of you. Once you graph the parabola, you simply take the intersections of the graph and x-axis as your x-intercepts. Therefore, it is a method.

b) Factoring

Factoring can be used to find x-intercepts, and after factoring, you can set each factor to 0 and solve for x, which are the x-intercepts. This method is also called the Zero Product Property.

c) Quadratic Formula

The quadratic formula is meant to solve for the x-intercepts, as the formula uses values of the coefficients to isolate x and solve for it.

Standard form of a Quadratic: [tex]ax^2 + bx + c = 0[/tex]

Quadratic Formula: [tex]x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}[/tex]

Plugging in the values of a, b, and c give you the solutions for x.

d) Completing the Square

Completing the square CAN be used to find the x-intercepts, but not directly so. It is used to convert standard form into a different form, known as the vertex form. It is meant so find the vertex, not the x-intercepts.

Therefore, Completing the Square is NOT a method to find the x-intercepts.

The number and types of solutions depend on the discriminant of the quadratic (the part under the square root in the QF).

Types of roots:

That means that a quadratic equation can have 0, 1, or 2 roots.

Answer:

Completing the square and 2, 1, and 0.

Step-by-step explanation:

Person above is correct!!

Check the picture below, so the parabola looks more or less like so, with the vertex half-way from the focus point and the directrix.

Now, the distance from the directrix to the focus point is only 1 unit, so half that, or namely the "p" distance is 1/2 unit, and since the parabola is opening downwards, "p" is negative, so

[tex]\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{cases} h=0\\ k=-\frac{1}{2}\\[1em] p=-\frac{1}{2} \end{cases}\implies 4\left( -\frac{1}{2} \right)\left( ~~y-\left( -\frac{1}{2} \right) ~~ \right)~~ = ~~(x-0)^2 \\\\\\ -2\left( y+\frac{1}{2} \right)=x^2\implies y+\cfrac{1}{2} =\cfrac{x^2}{-2}\implies y = -\cfrac{1}{2}x^2-\cfrac{1}{2}[/tex]

now, we could also write that as either

[tex]y = -\cfrac{1}{2}(x^2+1)\qquad or\qquad y = \cfrac{1}{2}(-x^2-1)[/tex]

Answer:

x = -9, y = 8

or,

(-9, 8)

Explanation:

Equation 1: 2x + 3y = 6

Equation 2: 4x - y = -44

To solve by substitution, make y subject for equation 2.

4x - y = -44

-y = -44 - 4x

y = 4x + 44

Substitute this y value into equation 1.

2x + 3(4x + 44) = 6

2x + 12x + 132 = 6

14x = 6 - 132

14x = -126

x = -9

Now, find y value:

y = 4x + 44

y = 4(-9) + 44

y = 8

To check for solution, insert x and y value in either equations.

2x + 3y = 6

insert values

2(-9) + 3(8) = 6

6 = 6

Hence, the solution is correct as both sides equal.

Therefore x = -9 and y = 8.

Substitution method

Equation (1) ——— 2x + 3y = 6

Equation (2) ——— 4x - y = -44

When using substitution method , make x the subject of equation (1).

2x + 3y = 6

2x = 6 - 3y

x = 3 - 3y/2

Substitute the value of x in equation (2) to find y.

4(3 - 3y/2) - y = -44

12 - 6y - y = -44

Subtract 12 from both sides.

12 - 12 - 7y = -44 - 12

-7y = -56

y = 8

Substitute the value of y in equation (1) to find x.

2x + 3y = 6

2x + 3(8) = 6

2x +24 = 6

Subtract 24 from both sides.

2x + 24 - 24= 6 - 24

2x = -18

x = -9

Therefore x = -9 and y = 8.

To check out the solution.

Equation (1) 2x + 3y = 6 , x = -9 , find y.

Substitute the value of x

2(-9) + 3y = 6

-18 + 3y = 6

Add 18 to both sides.

-18 + 18 + 3y = 6 + 18

3y = 24

y = 8

Substitute the value of y in equation (1) .

2x +3(8) = 6

2x + 24 = 6

Subtract 24 from both sides.

2x + 24 - 24 = 6 -24

2x = -18

x = -9

As you can see when you substitute the value of x and y in the equation(1) , you get the same value of x and y ,which is -9 and 8 respectively. So the solution is correct.

The area of the shaded region is B. 52 square inches.

It should be noted that the shaded region is in the shape of a rectangle.

Length = 18 - 4 - 1= 13 inches

Width = 6 - 2 = 4 inches

The area is:

= Length × Width

= 13 × 4

= 52 inches²

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Answer: this is the solutions

Solving a linear equation, we conclude that Margo's paycheck is of 96 dollars.

We know that she spends one quarter on a dress, one eight on shoes, and $24 on a birthday present.

We also know that she spent a total of $60.

If we define P as Margo's paycheck, then we know that:

(1/4)*P + (1/8)*P + $24 = $60

Now we just need to solve that linear equation for P, we will get:

(2/8)*P + (1/8)*P = $60 - $24 = $36

(3/8)*P = $36

P = (8/3)*$36 = $96

In this way, we conclude that Margo's paycheck is of 96 dollars.

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Answer: 98

Step-by-step explanation: 180 - 82 = 98

180 - 80 = 100, and 100 - 2 is 98.

The first step in the construction of a line perpendicular to line AB and passing through an external point m is:

a. place the compass needle on m and keeping the compass width the same, draw two arcs that intersect the line AB

To construct a perpendicular line, a compass and a scale are used.

Consider a line AB and an external point M.

Step 1:

Place the compass needle on M and keeping the compass width the same, draw two arcs that intersect the line AB.

Step 2:

Name the two intersecting points on line AB as X and Y.

Draw the arcs from these two points to the opposite side of point M with the same compass width.

Step 3:

Name the intersecting point of two arcs as N.

Draw a line connecting the two points M and N using scale.

Step 4:

Using a protractor, measure the angle of the line MN with the line AB.

The angle must show 90°.

Thus, the line MN perpendicular to line AB and passing through an external point M is constructed.

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Answer:

a. place the compass needle on m and keeping the compass width the same, draw two arcs that intersect the line AB

Step-by-step explanation: