%35 off the sale price
Original price: $135
Sale: $108.99
Multiply the sale price by the percentage off in decimal form:
108.99 x (35/100) = $38.15 (Discount)
Subtract that amount to the sale price:
108.99-38.15 = $70.84 (final price)
17x =99-13y13x = -39+13ysolve this system of linear equations. separate the x and y values with a comma
The equations are given by,
[tex]\begin{gathered} 17x=99-13y\ldots\text{.}\mathrm{}(1) \\ 13x=-39+13y--(2) \end{gathered}[/tex]Add equations (1) and(2).
[tex]\begin{gathered} 17x+13x=99-13y-39+13y \\ 30x=60 \\ x=\frac{60}{30}=2 \end{gathered}[/tex]Substitute the value of x in equation(1).
[tex]\begin{gathered} 17X2=99-13y \\ 34=99-13y \\ 34-99=-13y \\ -65=-13y \\ \frac{-65}{-13}=y \\ 5=y\text{ } \\ So,\text{ the values are x=2 and y=5} \\ (x,y)=(2,5) \end{gathered}[/tex]mx+b for m=7, x=8, b=9
mx + b
when m = 7 x = 8 and b = 9
Substitution
(7)(8) + (9)
Simplification
56 + 9
Result
65
What is the probability that a randomly selected student will earn a 91 - 100 given they study for two to three hours?
Explanation
Let the probability that those who earn 91-100 be Pr(A) and the probability that those who studied for 2-3 hours be Pr(B)
[tex]Pr(A|B)=\frac{Pr(A\cap B)}{Pr(B)}=\frac{22}{45}[/tex]Answer: 22/45 or 49.9%
aPerform the indicated operations and reduce to lowest terms. Assume that no denominator has a valueof zero.2x² x -2x2 + x - 21 8x + 16x² + 2xx - 3AnswerKeypadKeyboard Shortcuts
Given:
[tex]\frac{2x^2+x-21}{x^2+2x}\times\frac{8x+16}{x-3}[/tex]To find: The reduced form
Explanation:
Let us write it as the factored form.
[tex]\begin{gathered} \frac{2x^2+x-21}{x^2+2x}\times\frac{8x+16}{x-3}=\frac{2x^2+7x-6x-21}{x(x^{}+2)}\times\frac{8(x+2)}{x-3} \\ =\frac{x(2x^{}+7)-3(2x+7)}{x(x^{}+2)}\times\frac{8(x+2)}{x-3} \\ =\frac{(x-3)(2x+7)}{x(x^{}+2)}\times\frac{8(x+2)}{x-3} \end{gathered}[/tex]Simplifying we get,
[tex]\frac{8(2x+7)}{x}[/tex]Final answer: The simplest form is,
[tex]\frac{8(2x+7)}{x}[/tex]how do you know that a solution will be an imaginary number
A second degree equation of the form
[tex]ax^2+bx+c=0[/tex]always have solutions given by
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]this is called the general formula. In the general formula we have the number
[tex]b^2-4ac[/tex]inside the squared root, this number is called the discriminant of the equation and it plays an important role in the type of solutions we'll get.
• if the discriminant is greater than zero then the equation has two real solutions.
,• If the discrimiant is equal to zero then the equation has a real solution with multiplicity two.
,• If the discriminant is less than zero then the equation has two complex solutions, one conjugate of the other.
For example, think of the equation
[tex]x^2=-9[/tex]This, in standard form, is written as:
[tex]x^2+9=0[/tex]from here we see that a=1, b=0 and c=9. Then,
[tex]\begin{gathered} x=\frac{-0\pm\sqrt[]{0^2-4(1)(9)}}{2(1)} \\ =\frac{\pm\sqrt[]{-36}}{2} \\ =\frac{\pm\sqrt[]{-9}\sqrt[]{4}}{2} \\ =\frac{\pm2\cdot3i}{2} \\ =\pm3i \end{gathered}[/tex]Notice that in this case the discriminant was less than zero, then we will expect the solutions to be complex numbers.
A shipping container will be used to transport several 150-kilogram crates across thecountry by rail. The greatest weight that can be loaded into the container is 26000kilograms. Other shipments weighing 12900 kilograms have already been loaded intothe container. What is the greatest number of 150-kilogram crates that can be loadedinto the shipping container?
The problem says that the container can be loaded with a maximun weight of 26000 kg.
Also, other shipments weighing 12900 kg have already been loaded into the container, which means you need to subtract this to the maximum weight in order to know what is the actual capacity of the container, then:
[tex]26000\operatorname{kg}-12900\operatorname{kg}=13100\operatorname{kg}[/tex]Each crate weighs 150 kg, and you need to know what is the greatest number of crates that can be loaded, thus, you have to divide the capacity available into the weight of the crates, thus:
[tex]\frac{13100\operatorname{kg}}{150\text{ kg/crate}}=87.33\text{ crates}[/tex]Then, the greatest number of 150-kg crates that can be loaded into the shipping container is 87.
a market research company has conducted a survey to find the number of lcd television consumers have bought from a retail chain over five years. the table shows from the survey. estimate the average rate of lcd television sold by the retail chain between 2006 and 2008
In a table is given the number of lcd television consumers have bought from a retail chain over five years. We are asked to estimate the average rate of lcd television sold by the retail chain between 2006 and 2008.
To estimate this average we will consider only the following three numbers
[tex]73000,97000,92000[/tex]They correspond to the Tv's sold in 2006,2007 and 2008 respectively , we use the following formula for the average rate
[tex]\text{ Average rate of sells}=\frac{Difference\text{ between number of Tv sold}}{Time\text{ transcurred}}=\frac{92000-73000}{2008-2006}[/tex]Making the calculations we find that:
[tex]\text{ Average rate of sells=}\frac{19000}{2}=9500[/tex]Looking at the options for this question, we found that the correct one is the option C
The quotient of y and three is negative nine. how do I write a equation for this sentence and how do I solve it show the work please ^-^
a. By definition, the Quotient is the result of a Division.
A Division can be written as a fraction:
[tex]\frac{a}{b}[/tex]Where "a" is the numerator and "b" is the denominator.
You also need to remember that word "is" indicates that you must use:
[tex]=[/tex]Knowing the explained above, you can set up the following equation using the information given in the exercise:
[tex]\frac{y}{3}=-9[/tex]b. You need to solve for "y" in order to find its value. You can apply the Multiplication property of equality by multiplying both sides of the equation by 3. Then:
[tex]\begin{gathered} (\frac{y}{3})(3)=(-9)(3) \\ \\ y=-27 \end{gathered}[/tex]The answers are:
a.
[tex]\frac{y}{3}=-9[/tex]b.
[tex]y=-27[/tex]O COUNTING AND PROBABILITY Introduction to permutations and combinations Suppose we want to choose 2 colors, without replacement, from the 5 colors red, blue, green, purple, and yellow. (a) How many ways can this be done, if the order of the choices is taken into consideration? X Х ? (b) How many ways can this be done, if the order of the choices is not taken into consideration?
Answer: We are given five colors, Red Blue Green Purple and Yellow, We need to place them into two, and would like to know the total combinations, A The order matters, B Order does not Matter.
A-The order matters:
[tex]\begin{gathered} \text{slots =2 } \\ \text{Colors = =5} \end{gathered}[/tex]Therefore we have:
[tex]5\cdot4\cdot3\cdot2\cdot1=60\cdot2\cdot1=120\text{ ways}\rightarrow\text{ Because order mattered}[/tex]B-The order does not matter:
[tex]\frac{5!}{2!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=60\rightarrow Because\text{ the order does not matter}[/tex]Note!, When the order does matter, we are counting each possibility twice, and when it does not, we only count once.
whaf does the notation f(2) mean?
Explanation:
f(x) is a notation of a function x
E.g f(x) = ax + b
This implies that the equation is a function of x
F(2) means the value of the output when x = 2
Therefore, f(2) means the output (y - value) whe x = 2
what is the area of the pentagon with a radius of 24 mm?
1360.01 square mm
Explanations:
The formula for calculating the area of a regular pentagon is expressed as:
[tex]A=\frac{5}{2}r^2sin72^0[/tex]where
r is the radius of the pentagon
Given the following parameters
r = 24mm
Substitute
[tex]\begin{gathered} A=\frac{5}{2}(24)^2sin72^0 \\ A=\frac{5}{2}\times572sin72^0 \\ A=2.5\times544.00 \\ A=1360.01mm^2 \end{gathered}[/tex]Hence the area of the regular pentagon is 1360.01 square mm
One spring day, Jaya noted the time of day and the temperature, in degrees Fahrenheit. Her findings are as follows: At 6 a.m., the temperature was 50° F. For the next 4 hours, the temperature rose 1° every 2 hours. For the next 3 hours, it rose 3° per hour. The temperature then stayed steady until 6 p.m. For the next 4 hours, the temperature dropped 3° every 2 hours. The temperature then dropped steadily until the temperature was 54° at midnight. On the set of axes below, graph Jaya's data.
We were given the following:
Time-Temperature data
[tex]\begin{gathered} 6am=50\degree F \\ \text{Over the next 4 hours, temperature rose }1\degree\text{every 2 hours:} \\ 7am=50.5\degree F \\ 8am=51\degree F \\ 9am=51.5\degree F \\ 10am=52\degree F \\ \text{Over the next 3 hours, temperature rose }3\degree\text{every 2 hours:} \\ 11am=55\degree F \\ 12pm=58\degree F \\ 1pm=61\degree F \\ \text{The temperature stays steady until 6pm:} \\ 2pm=61\degree F \\ 3pm=61\degree F \\ 4pm=61\degree F \\ 5pm=61\degree F \\ 6pm=61\degree F \end{gathered}[/tex]For the next 4 hours, the temperature dropped 3° every 2 hours. We have:
[tex]\begin{gathered} 7pm=59.5\degree F \\ 8pm=58\degree F \\ 9pm=56.5\degree F \\ 10pm=55\degree F \end{gathered}[/tex]The temperature then dropped steadily until the temperature was 54° at midnight. We have:
[tex]\begin{gathered} 11pm=54.5\degree F \\ 12am=54\degree F \end{gathered}[/tex]We will now proceed to plot these data points on a graph. We have:
Find the length of the third side. 20 28 A) 19.6 B) 384 C) 8 D) 48
ANSWER
A) 19.6
EXPLANATION
This is a right triangle, so we have to use the Pythagorean theorem to find the third side's length.
First we have to identify the hypotenuse of the triangle. The hypotenuse is always the opposite side to the right angle. In this case, the hypotenuse is 28 units long. Therefore, what we want to find is the length of one of the legs.
Let 'x' be the missing length:
[tex]28^2=x^2+20^2[/tex]This is the pythagorean formula for this triangle. Solving for x:
[tex]\begin{gathered} x=\sqrt[]{28^2-20^2} \\ x=\sqrt[]{784-400} \\ x=\sqrt[]{384} \\ x=8\sqrt[]{6} \\ x\approx19.6 \end{gathered}[/tex]The third side's length is 19.6 units (rounded to the nearest tenth).
Given Rectangle MNPQ, what is the length of MQ? (Hint: Draw a perpendicular bisector & use Pythagorean Theorem. Round your answer to the nearest tenth!)
In order to find the length of MQ, we can first draw a segment from the intersection point of the diagonals (let's call it D) and parallel to sides NM and PQ, going down to point K, like this:
The segment DK has a length of half the length of PQ, so we have DK = 2.
Now, we can use the Pythagorean Theorem to find the length of KQ:
[tex]\begin{gathered} DQ^2=DK^2+KQ^2 \\ 5^2=2^2+KQ^2 \\ 25=4+KQ^2^{} \\ KQ^2=25-4 \\ KQ^2=21 \\ KQ=4.58 \end{gathered}[/tex]The length of KQ is half the length of MQ, so we have:
[tex]\begin{gathered} KQ=\frac{MQ}{2} \\ MQ=2\cdot KQ \\ MQ=2\cdot4.58 \\ MQ=9.16 \end{gathered}[/tex]Rounding to the nearest tenth, we have MQ = 9.2 units
1.34 x-67.8 =I need help
-90.852
Explanation
1.34*-67.8
Step 1
count the number of digits after the dot in each number
[tex]\begin{gathered} 1.34\text{ }\Rightarrow\text{ 2 digits after the dot} \\ 67.8\Rightarrow1\text{ digit} \end{gathered}[/tex]total = 2+1= 3 digits after the points.
Step 2
eliminate the dots and do the multiplication
[tex]\begin{gathered} 134\cdot-678 \\ +\text{ multiplied by - is =-, then} \\ 134\cdot-678=-90852 \end{gathered}[/tex]Step 3
go to the last number of the result(rigth side) and move those e spaces(the total count of digits in step 1) from right to left, and mark the point) it is
[tex]\begin{gathered} \text{number of spaces to move= 3} \\ \text{result= -90852} \\ \text{then} \\ -90.852 \end{gathered}[/tex]so, the answer is -90.852
I would like to get help on this question please! And how to solve it
We are given the following product
[tex](1\text{ -x\rparen}\cdot(x+3)\cdot(7x+3)^2[/tex]we want to find the degree, leading coefficient and constant coefficient of this product. To do so, we should find first this quantities from each of the polynomials.
Recall that the degree of the polynomial is the highest power of the variable (in this case x). Recall that the leading coefficient is the coefficient that is multiplying the power of x with the highest degree. Finally, recall that the constant coefficient of the polynomial is the coefficient that has no variable.
So we have
[tex](1\text{ -x\rparen}[/tex]in this case, the highest power of x is 1. So the degree of this polynomial is 1. In this case, the coefficient that multiplies the highest power of x is -1. So the leading coefficient is -1. Finally, the constant coefficient is 1.
For the other polynomial we have
[tex](x+3)[/tex]in this case the degree is 1, the leading coefficient is 1 and and the constant coefficient is 3.
Finally, for the other polynomial we have
[tex](7x+3)^2=49x^2+42x+9[/tex]so for this polynomial, the degree is 2, the leading coefficient is 49 and the constant coefficient is 9.
Now, we simply use this information to calculate it for the whole product.
The degree of the polynomial is the sum of all degrees. So the degree of the polynomial is 1+1+2=4.
The leading coefficient is the product of the leading coefficients. So the leading coefficient is -1*1*49=-49.
The constant coefficient is the product of the constant coefficients. So the constant coefficient is 1*3*9=27.
Add 63∠50 and 21∠-62Question 16 options:(30.64, 66.84)(50.36, 66.84)(50.36, 29.72)(84, -12)
Given the expressions:
[tex]63\angle50\text{ and 21}\angle-62[/tex]Let's add the expressions.
To add the expression, let's seperate the real part and the part with the angle and add seperately.
We have:
Real part: 63 + 21 = 84
Angle: ∠50 + ∠-62 = -12
Therefore, the answer after adding both expressions is:
(84, -12)
ANSWER:
(84, -12)
2×3+42×2what's 2 * 3 + 4 * 2
To know the result, we need to make first the multiplications and then the sum, So:
2 * 3 + 4 * 2
( 2 * 3) + (4 * 2)
6 + 8
14
For the following expression, the result is:
2×3 + 42×2
(2×3) + (42×2)
6 + 84
90
Answer: 14
In the following problems, find (a) the compound amount and (b) the compound interest for the given investment and annual rate. 1. $4,000 for 7 years at 6% compounded annually. 2. $5,000 for 20 years at 5% compounded annually. 3. $700 for 15 years at 7% compounded semiannually.
Answer:
Step-by-step explanation:
The compound interest is represented by the following equation:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where}, \\ A=\text{ compound interest} \\ P=\text{ principal} \\ r=\text{ interest rate} \\ t=\text{time} \\ n=\text{times per year compounded} \end{gathered}[/tex]Therefore, for:
1. P=$4,000, t=7 years at r=6%
[tex]\begin{gathered} A=4000(1+0.06)^7 \\ A=\text{ \$6,014.52} \\ \\ \text{For the interest, subtract the principal amount:} \\ I=\text{ \$6,014.52}-\text{ \$4,000} \\ I=\text{ \$2,014.52} \end{gathered}[/tex]2. P= $5,000, t=20 years at r=5%
[tex]\begin{gathered} A=5000(1+0.05)^{20} \\ A=\text{ \$13,266.48} \\ \\ \text{For the interest, subtract the principal amount:} \\ I=\text{ \$13,266.48-\$5,000} \\ I=\text{ \$8,266.48} \end{gathered}[/tex]3. P=$700, t=15 years at 7%, n=2
[tex]\begin{gathered} A=700(1+\frac{0.07}{2})^15\cdot2 \\ A=1964.75 \\ \\ I=1964.75-700 \\ I=1264.75 \end{gathered}[/tex]3) Solve the system of equations below by using ELIMINATION and record your solution. 2x + 3y = 15 X-3y = 3 X= y =
The equations are 2x + 3y = 15 and x - 3y = 3.
Add the equations to eliminate the y term and obtain the value of x.
[tex]\begin{gathered} 2x+3y+x-3y=15+3 \\ 3x=18 \\ x=\frac{18}{3} \\ =6 \end{gathered}[/tex]Substitute 6 for x in equation 2x + 3y = 15 to obtain the value of y.
[tex]\begin{gathered} 2\cdot6+3y=15 \\ 3y=15-12 \\ y=\frac{3}{3} \\ =1 \end{gathered}[/tex]Thus solution of equations is x = 6 and y = 1.
Find the coordinates of the point on a circle with radius 16 corresponding to an angle of \frac{5\Pi}{9}. If your answer is not an integer then round it to the nearest hundredth. The x component of the coordinate is AnswerThe y component of the coordinate is Answer
Answer:
The coordinate is (15.99, 0.49).
Explanation:
Given that:
Radius of the circle = 16
The x component of the coordinate is
[tex]\begin{gathered} rcos\theta=16\cdot cos\lparen\frac{5\pi}{9}) \\ =15.99 \end{gathered}[/tex]The y component of the coordinate is
[tex]\begin{gathered} rsin\theta=16sin\left(\frac{5\pi}{9}\right) \\ =0.49 \end{gathered}[/tex]The coordinate is (15.99, 0.49).
20. How many ways can we select five door prizes from eight different ones (meaning each prize is unique) and distribute them among five people?
Since five door prizes are to be selected from eight different ones, we have
[tex]C^{8_{}}_5[/tex]Recall that
[tex]C^n_r=\frac{n!}{(n-r)!r!}[/tex]Thus,
[tex]C^{8_{}}_5=\frac{8!}{(8-5)!5!}=\frac{8!}{3!\times5!}[/tex]This gives
[tex]\frac{8\times7\times6\times5!}{3\times2\times1\times5!}=56[/tex]To be distributed among five people results to
[tex]P^{56}_5=\text{ }\frac{56!}{(56-5)!}=458377920[/tex]
How much money will you need to invest initially to have $750.00 in 10 years and 8 months if the money is compounded daily at an annual rate of 2 1/2%?A.$585.90B. $574.45C. $576.33D. $574.60
Given:
The amount after 10 years and 8 months, A=$750.00.
The rate of interest, r =2 1/2 %.
The period of time, t =10 years and 8 months.
The interest is compounded daily
Required:
We need to find the intial investment amount.
Explanation:
Conver the period of time to years.
[tex]1\text{ year =12 months.}[/tex][tex]\frac{8}{12}\text{ year =8 months.}[/tex][tex]10\text{ +}\frac{8}{12}\text{ years =10 years and 8 months.}[/tex][tex]10\text{ +}\frac{8}{12}\text{ years =10 years and 8 months.}[/tex][tex]10\text{ }\times\frac{12}{12}\text{+}\frac{8}{12}\text{ years =10 years and 8 months.}[/tex][tex]\frac{120}{12}\text{+}\frac{8}{12}\text{ years =10 years and 8 months.}[/tex][tex]\frac{120+8}{12}\text{ years =10 years and 8 months.}[/tex][tex]\frac{128}{12}\text{ years =10 years and 8 months.}[/tex]We get t =128/12.
The annual interest rate is
[tex]r=2.5\text{ \%.}[/tex][tex]r=0.025.[/tex]The number of days in a year = 365 days.
The money is compounded daily, n=365.
Consider the formula to find the amount in compound interest.
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Substitute A =750, r=0.025, n=365 and t =128/12 in the formula.
[tex]750=P(1+\frac{0.025}{365})^{365\times\frac{128}{12}}[/tex][tex]750=P(1+\frac{0.025}{365})^{3893.333}[/tex][tex]P=\frac{750}{(1+\frac{0.025}{365})^{3893.333}}[/tex][tex]P=574.4516[/tex]Final answer:
The initial amount is $ 574.45.
Give the slope of a parallel line from the given equation. y=4/5x-3
Parallel lines are equal and have the same slope. The slope law for parallel lines can be expressed as follows
[tex]\begin{gathered} m_1=m_2 \\ \text{where} \\ m_1=\text{slope 1} \\ m_2=\text{slope 2} \end{gathered}[/tex]The slope point equation can be express as below
[tex]\begin{gathered} y=mx+c \\ \text{where} \\ m=\text{slope or gradient} \\ c=\text{intercept} \\ \text{Therefore} \\ y=\frac{4}{5}x-3 \\ \text{the slope will be }\frac{4}{5} \end{gathered}[/tex]The slope is 4/5
If 0° ≤ A ≤ 180°, find ∠A, to the nearest tenth of a degree.
a) A = 140.6 degrees
b) A = 16.6 degrees
c) C = 122.9 degrees
d) C = 64.6 degrees
Explanation:[tex]\begin{gathered} \cos A=-0.7732 \\ \\ A=\cos^{-1}(-0.7732)=140.6^o \end{gathered}[/tex][tex]\begin{gathered} \sin A=0.2853 \\ \\ A=\sin^{-1}(0.2853)=16.6^o \end{gathered}[/tex][tex]\begin{gathered} \tan C=-1.5477 \\ \\ C=\tan^{-1}(-1.5477)=180-57.1327=122.9^o \end{gathered}[/tex][tex]\begin{gathered} \cos C=0.4288 \\ \\ C=\cos^{-1}(0.4288)=64.6^o \end{gathered}[/tex]Find the value of cos(2x) if sin(x) = 5/13 and x is in Quadrant II.
The given equation is:
[tex]\sin(x)=\frac{5}{13}[/tex]It is required to find the value of cos(2x) if x is in the second quadrant.
Recall the identity:
[tex]\cos(2x)=1-2\sin^2(x)[/tex]Substitute sin(x)=5/13 into the identity:
[tex]\cos(2x)=1-2\left(\frac{5}{13}\right)^2=1-2\left(\frac{25}{169}\right)=1-\left(\frac{50}{169}\right)=\frac{119}{169}[/tex]The answer is 119/169.
In rectangle ABCD, diagonal AC=x+10 and diagonal BD=2x-30. Find the value of x.
Does this graph represent a function? Why or why not?2214010A. Yes, because it passes the vertical line test.O B. Yes, because it is a curved line.C. No, because it fails the vertical line test.D. No, because it is not a straight line.
We are given the following graph of a function and asked to select the correct answer:
Our answer is: Yes, this is the graph of a function, because it passes the "vertical line" test.
So we select option A on the list od possible answers
Determine if there enough information to prove that each pair of triangles is congruent by SSS,SAS, or ASA. write the congruence statements represented by the markers in each diagram.ΔPRT ≅ ΔTVP
Triangle PRT =~ Triangle TVP
can be proved congruent with SAS
because there are two line sides similar
PR=VT = 12
and
PV= RT = 5
And there are 2 similar angles
PRINCIPLE : 2 lines are parallel if ,internal alternate angles are similar, or congruents
How many different 6-letter words can be madea. if the first letter must be C, X, Q, or M and no letter may be repeated?b. if repeats are allowed (but the first letter is C, X, Q, or M)?c. How many of the 6-letter words (starting with C, X, Q, or M) with no repeats endin R?a. If the first letter must be C, X, Q, or M, and all the letters in the word must be different, then there are6-letter words.
First let;s solve part (a)
Now first place can be filled in 4 ways
Second place can be filled in 22 ways
Third place can be filled in 21 ways
Fourth place can be filled in 20 ways
Fifth place can be filled in 19 ways
Sixth place can be filled in 18 ways
So total number of 6 letter words will be
[tex]\begin{gathered} 4\times22\times21\times20\times19\times18 \\ =12,640,320 \end{gathered}[/tex](b)
If digits are can be repeated then first pace can be filled in 4 ways but trest of the positions can be filled in 26 way s
So total numbe rof words will be
[tex]\begin{gathered} 4\times26\times26\times26\times26\times26 \\ =47,525,504 \end{gathered}[/tex]c) Now first place can be filled in 4 ways
And sixth place can be fille din 1 way only
and no letter can be repeated
So second place can be filled in 21 ways
Third place can be filled in 20 ways
Fourth place can be filled in 19 ways
Fifht place can be filled in 19 ways
So total numbe of words will be
[tex]\begin{gathered} 4\times21\times20\times19\times18\times1 \\ =574,560 \end{gathered}[/tex]The last part is same as the part (a)