Which expression is equivalent to (2 Superscript one-half Baseline times 2 Superscript three-fourths Baseline) squared?
RootIndex 4 StartRoot 2 cubed EndRoot
StartRoot 2 Superscript 5 EndRoot
RootIndex 4 StartRoot 4 cubed EndRoot
StartRoot 4 Superscript 5 EndRoot

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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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:

None of the answers

Step-by-step explanation:

There are 60 minutes in one hour so to find the number of minutes in two hrs we multiply 2 by 60: 2*60 = 120 minutes

We then add the extra 20 minutes: 120+20 = 140 minutes

None of the answers include 140 minutes.

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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The researcher is using a within-subjects factorial design.

According to statement

A researcher conducts an experiment to determine whether the type of car and color of the car impact how fast people drive.

This can be done by within-subjects factorial design.

In a within-subjects factorial design, all of the independent variables are manipulated within subjects. This would mean that each participant was tested in all conditions.

Another common example of a within-subjects design is medical testing.

So, The researcher is using a within-subjects factorial design.

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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 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:

Answer:

Step-by-step explanation:

Cylinder volume formula:

Given:

The first cylinder, has base radius - r and height - h. Its volume is:

The second cylinder has:

Its volume is:

Substitute the value of the first cylinder:

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:

76.25

Step-by-step explanation:

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.

Answer: this is the solutions

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]

Mean = $ 4. 72

Median = $ 4. 5

Mode = No mode

MAD = $ 4. 8

IQR = $ 0. 5

Measure of center = $ 4. 5

Measure of variability = $ 2. 50

Given the data;

$3.50, $4.25, $6, $4.75, $5.80, $4

Mean = sum of data/ number of data values

Mean = [tex]\frac{3. 50 + 4. 25 + 6 + 4. 75 + 5. 80 + 4}{6}[/tex]

Mean = $ 4. 72

Median is the number in the center when the data is arranged in an ascending order

$3. 50, $4, $4. 25, $4. 75, $5. 80, $6

Median = [tex]\frac{4. 25 + 4. 75}{2}[/tex]

Median = $ 4. 5

Mode is the number with the highest repeats.

From the given data, none of the numbers was repeated.

Thus, there is no mode.

MAD, mean absolute deviation = |data value - mean|

MAD = [tex]|(3. 50 - 4. 72) + (4. 25 - 4. 72) + (6 - 4. 72) + (4. 75 - 4. 72) + (5. 80 - 4.72) + (4- 4. 72) |[/tex]

MAD = [tex]1. 22 + 0. 47 + 1. 28 + 0. 03 + 1. 08 + 0. 72[/tex]

MAD = $ 4. 8

IQR, interquartile range = Quartile 3 - quartile 1

IQR = 4. 75 - 4. 25

IQR = $0. 5

Note that  the mean and median are the most common measures of center.

A measure of variability is a single number that is  used to describe the spread of a data set

The measure of the center best for the data is the median = $4. 5

Measure of variability = range = highest value - lowest value = $6 - $3. 50 = $2. 50

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Answer:

Yes. If the squares of the 2 short sides add up to the square of the hypotenuse, the triangle contains a right angle. Hope this helped.

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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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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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  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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Nguyen saves 15% of his income

The given parameters are:

Savings = $120

Income = $800

The percentage saved is calculated as:

Percentage = Savings/Income * 100%

This gives

Percentage = 120/800 * 100%

Evaluate

Percentage = 15%

Hence, Nguyen saves 15% of his income

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Answer:

d=(x^2-x^1)+(y^2-y^1)^2

Answer:

[tex]\mathfrak{Distance \: between \: two \:points}[/tex]

The distance between two points [tex]A(x_1,y_1)[/tex] and [tex]B(x_2,y_2)[/tex] is given by the formula,

[tex]AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2 }[/tex]

Proof: Let X'OX and YOY' be the x-axis and y-axis respectively. Then, O is the origin.

Let[tex]A(x_1,y_1) [/tex] and [tex]B(x_2,y_2)[/tex]

be the given points.

Draw AL perpendicular to OX, BM perpendicular to OX and AN perpendicular to BM

Now,[tex]OL=x_1,OM=x_2,AL=y_1 \: and \: BM=y_2[/tex]

[tex] \therefore{AN=LM=(OM-OL)=(x_2-x_1)}[/tex]

[tex] \: \: \: BN=(BM-NM)=(BM-AL)=(y_2-y_1)[/tex]

In right angled triangle [tex] \triangle \: ANB[/tex],by Pythagorean theorem,

We have,

[tex] \: \: \: AB^2=AN^2+BN^2[/tex]

[tex]or,AB^2=(x_2-x_1)^2+(y_2-y_1)^2[/tex]

[tex] \therefore \: AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]

Thus ,the distance between the points A(x_1,y_1) and B(x_2,y_2) is given by,

[tex] \implies \boxed{AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }[/tex]

Answer:

H. 10 h

Step-by-step explanation:

40.50-5.5 = 35

35/3.5 = 10 hours

The number of credit hours a student have to take for the two tuition costs to be equal is 2.5 hours. Option A

c(h) = 250 + 200h

s(h) = 300 + 180h

Where,

To have equal tuition in both schools,

equate both function

250 + 200h = 300 + 180h

200h - 180h = 300 - 250

20h = 50

h = 50/20

h = 2.5 hours.

The number of credit hours a student have to take for the two tuition costs to be equal is 2.5 hours.

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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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The rotation about 90 degrees clockwise from the origin would give H'( -2, 3) , I' (1 , -1) and J' (-4, 0)

Note that rotation about in clockwise 90 degrees direction when M(h , k) rotated about the origin, the new rotation coordinates would be in the form M' (k , -h)

Coordinates of the preimage were given as;

H ( -3, -2)

I ( 1, 1 )

J ( 0, -4)

The coordinates after a 90 degrees rotation about the origin, would be

For H ( -3, -2)

h = - (-3) = 3

k = -2

H' = ( -2, 3)

For I ( 1, 1)

h = - (1) = -1

k = 1

I' = ( 1, -1)

For J ( 0, -4)

h = - (0) = 0

k = -4

J' = ( -4, 0 )

Thus, the rotation about 90 degrees clockwise from the origin would give H'( -2, 3) , I' (1 , -1) and J' (-4, 0)

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The concept of factoring quadratic equations simply is one of the ways to solve for the zeroes of the quadratic equation.

The concept of factoring quadratics is a technique which is characterized by expressing the quadratic equation ax2 + bx + c = 0 as a product of its associated linear factors as (x - f)(x - g), in which case, f and g are the roots of the quadratic equation ax2 + bx + c = 0.

Factoring is an important concept that helps express equations of any type in simpler forms The mode of operation of the factoring works such that, the polynomial is rewritten in a simpler form and consequently, expressed as a product of it's linear factors.

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We have proven that ∠BEC ≅ ∠BCE by using the alternate interior angles theorem

From the question, we are to write a proof to prove that ∠BEC ≅ ∠BCE

From the given information, we have that

BC || EF

Note that line DE is a transversal

Thus, by the alternate interior angles theorem, we have that

∠BCE ≅ ∠FEC

Also, we were given that line EC bisects ∠BEF

That is,

Line EC divides ∠BEF into two equal angles.

Therefore,

∠BEC ≅ ∠FEC

By the substitution property of equality, we get

∠BEC ≅ ∠BCE

Hence, we have proven that ∠BEC ≅ ∠BCE by using the alternate interior angles theorem

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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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