Algebra Questions

Consequently the DE 3 is actually the following system of ODE in the functions t X t t X t t X t d dX dk X 1 X 5 X dik dt dk X dik 12 1 X 4 dk X di di 1 d x 1 1 x dx d dX dik 1 X 1 dX dx dt d X dik t At this stage we become more specific about the features of the ODE 3 or equivalently about the system 4 Let I be an open internal and let denote an open subset of R We consider the open sets IxNxR x xR there being k 1 copies of R in the above Cartesian product This set is being designated to accommodate the variable quantities dk X di 1 dX dt 4 Clearly the function f appearing on the right hand side of 3 must have this set as its domain of definition Choosing 1 1 x En and w w win R we form the initial condition fo W W w Now the initial value problem for the ODE 3 is the following pair dk X di 1 ff x ToX W W 1 5 By a solution of the initial value problem 5 we mean an at least k mously diffamanti 01

we regard the equation 3 as standard form of an UDE OI course the ODE has order k Note that the function X tX t is a vector valued function of the real variable t and as such it is a curve in R Each X has n components X t X 1 X 1 X t and therefore all the derivative of it has n components X 1 X 1 for 1 k Consequently the function f appearing on the right hand side of 3 has n components f f ff each f being a real valued function d X 1 dt MATHEMATICS Consequently the DE 3 is actually the following system of ODE in the functions t X t t X t t X t d X dX 1 X d di d X dr 1 1 X d X di ft X t X t dX d X dt dit 1 dX dt dx dt 1 3 At this stage we become more specific about the features of the ODE 3 or equivalently about the system 4 Let I be an open internal and let denote an open subset of R We consider the open sets Ixnx R xXR there being k 1 copies of R in the above Cartesian product This set is being designated to accommodate the variable quantities 4 dit 1 Clearly the function f appearing on the right hand side of 3 must have this set as its domain of definition dX dk X Choosing 1 1 x En and www in R we form the initial Now the initial value problem for the ODE condition fo W WW 3 is the following pair d X f

To begin with we reorganize the form 1 of the ODE in the following manner Unraveling it we separate the top order derivative and express it as a function of the remain variable quantities namely dX d X t X 1 that is we dt di form the equation ft X dX d x dt di d X di We regard the equation 3 as the standard form of an ODE of course the ODE has order k Note that the function X t X t is a vector valued function of the real variable t and as such it is a curve in R Each X t has n components X t X 1 X 1 X 1 and therefore all the derivative of it has n components d X 1 X 1 X dt X for 1 k Consequently the function f appearing on the right hand side of 3 has n components f f ff each f being a real valued function MATHEMATICS Consequently the DE 3 is actually the following system of ODE in the functions t X t t X X t da X dik 1 ft X d X dik f 1 X d X dik dX dt f t X dx t X t dt 1 d X dik dX dt dx dX dt dit 1 d X di 1 At this stage we become more specific about the features of the ODE 3 or equivalently about the system 4 dit 1 3 3 dk X dik 1 Let I be an open internal and let denote an open subset of R We consider the open sets IxNxR x XR there being k 1 copies of R in the above Cartesian product This set is being designated to accommodate the variable quantities 4 Clearly the function f appearing on the right hand side of 3 must have this set as its domain of definition dX dk X To W W 1 Choosing t 1 x En and www in R we form the initial condition o W WW Now the initial value problem for the ODE 3 is the following pair 5

11 Graph the system of inequalities Show at least two points for each inequality Lesson 1 08 2x 3y 15 y 5 4x 3x 5y 10 E 10 8 6 4 2 10 8 CANNOT USE 6 4 2 246 8 10 2 and and and 4 777 6 8 10 12 The shaded region represents the number of pounds of peanuts x and pounds of pecans y that a company plans to use in their trail mix List three combinations of peanuts and pecans that they can use in their recipe and three combinations that they cannot use Lesson 1 09 CAN USE 422222 45 35 30 25 20 15 10 5 0 5 0 5 10 15 20 25 30 35 40 45 50 55 60

Proof Recall that both x and y being solutions of the initial value problem satisfy the integral equations on their domain intervals x 1 xo f s x s ds y t x ff s y s ds Therefore x t y t Sis s x s f s y s ds which implies x 1 y 1 0 s x s s y s 0 Kx s y s for all t te Applying Gronwalls result with A 0 we get 0 x 1 y 1 0 for all t t This gives the desired equality x t y t for all tJJ Towards the uniqueness of the solution of the initial value problem 3 we consider all the solutions of the initial value problem 3 Let the totality of them be denoted by x J XEA the solutions x being thus indexed by a suitable indexing set A Above we have verified that any two solutions say x and x are equal on the overlap of their domains Therefore we patch together all the solutions to get a maximal solution is the solution defined on the largest open interval It is obtained as follows Let J U J 2 A clearly Jis an open sub internal of I with XJ and all the solutions x patch up to get a solution x J of the initial value problem 3

Which description is correct for the polynomial 3x 4x 7 O cubic binomial O cubic trinomial O quartic trinomial O quadratic trinomial

What is the degree of the polynomial 5pq2 3p6 5p6q 6p q5 Enter your answer in the box

Simplify 5x 2x 5 3x 3x 1 O 2x 5x 4 O 2x 5x 4 O 8x 5x 4 O 2x x 6

Simplity 5x 2 3x 4x 3 O 12x 1 O 3x x 5 O 3x 9x 1 O 3x 9x 1

The monthly cost of a certain texting plan is given by the function y 0 07x 14 95 where y is in dollars and x is the number of texts sent in a month Find and interpret the slope and y interce the linear equation A m 14 95 The number of texts sent in a month increases 14 95 for every dollar spent b 0 07 The number of texts that can be sent when x 0 is 0 07 B m 0 07 The number of texts sent in a month increases 0 07 for every dollar spent b 14 95 The number of texts that can be sent when x 0 is 14 95 OC m 14 95 The cost of the texting plan increases 14 95 for every text sent b 0 07 The cost of the texting plan is 0 07 if no texts are sent for the month O D m 0 07 The cost of the texting plan increases 0 07 for every text sent b 14 95 The cost of the texting plan is 14 95 if no texts are sent for the month C

D Graph the linear function using the slope and the y intercept y 6x 3 Use the graphing tool to graph the linear equation Use the slope and the y intercept when drawing the line Click to enlarge graph 12 10 12 10 18 6 Ay 48 10 10 42

Solve triangle ABC If an answer does not exist enter DNE Round your answers to one decimal place Below enter your answers so that A is smaller than A b 127 c 163 LB 49 ZA 55 4 2C 75 6 a 138 5 X X O O LA 26 6 4C 104 4 a2 75 4 XXX 0 0 O

What is the value of 22 2 0 O O 27 8 27 8 0 971 27 27

ork 7 1 Evaluate the function at the given value of the variable f x 3x 5x 3 a f 3 b f 4 a f 3 39 b f 4

We begin some more generalities related to first order ODE As in the preceding part I denoted an open interval and 2 an open subset of R We consider a function f Ixn R giving rise to the first order ODE dX dt f t x dX dt Note that for each 1 EI held fixed the map ft R x f 1 x is a vector field on Interpreting as the time variable we call the function 7 a time dependant vector field on n And often we call a solution X J of the ODE 8 an integral curve of the vector field f An initial condition for the ODE 8 consists of a pair too with to El x En and the associated initial value problem is f t X x t Xo X J N Finally recall that a solution of 9 is an at least once continuously differentiable curve J being an open interval with 1 EJ CI satisfying dX 1 dt 7 f t X t for all tEI and the initial condition X to x MATHEMATICS 8 In the context of the IVP 9 we consider yet another equation the following integral equation in an unknown function X J X 1 Xo f s x s ds tEJ 10 Proposition 1 Following result relates solutions of the IVP 9 and those of the integral equation 10 J s X s ds 9 X to x A continuously differentiable curve X J J being an open subinterval of I with 1 EJ is a solution of the IVP 9 if and only if it satisfies the integral equation 10 Proof I First suppose that the curve X J satisfies the integral equation 10 Putting 1 1 in 10 we get 0 X f s X s ds Thus X satisfies the initial condition Next differentiating 10 we get dX t d 0 f s X

Plot the ordered pair 5 9 on the rectangular coordinate plane Plot 5 9 Next question 10 10 8 8 2 4 6 8 Ay TO E G

Plot the point 0 4 in a rectangular coordinate system Plot 0 4 10 6 2 10 8 6 4 2 2 Ay 6

Plot the intercepts to graph the equation x y 8 Use the graphing tool to graph the equation Use the intercepts when drawing the line If only one intercept exists use it and another point to draw the line Click to enlarge graph O Points 0 of 1 Ay 10 8 6 2

What is a y intercept of a graph Choose the correct answer below OA A y intercept of a graph is the y coordinate of a point where the graph intersects the y axis OB A y intercept of a graph is the y coordinate of a point where the graph intersects the x axis OC A y intercept of a graph is the x coordinate of a point where the graph intersects the x axis OD A y intercept of a graph is the x coordinate of a point where the graph intersects the y axis

5 Find the equation for the line shown on the right What is the equation for the line shown on the right Points 0 Ay 10 10 8 84 2

Find the equation for the line shown on the right What is the equation for the line shown on the right 10 B Ay 10

Watch the video and then solve the problem given below Click here to watch the video Calculate the slope between 3 4 and 4 2 The slope is Type an integer or a fraction

1 Evaluate the function for the given value of x 3 ifx 0 2 if x 0 g x f x https goformative com formatives 6329af67d481e6e0db19dbcc 1 2 5 g 7 9 h 4 2 f 4 6 g 0 10 h 2 x 5 if x 3 2x 1 if x 3 3 f 0 7 g 1 11 h 1 h x zoom in 2x if x 2 4 8 g 3 12 h 6 N A

Complete the following proofs 11 Given AB is the bisector of 4CAD Prove x 9 Statements 1 AB is the bisector of ZCAD 2 ZCAB ZBAD 3 m CAB m BAD 4 7x 2 5 x 4 5 7x 2 5x 20 6 D 7 F Reasons 1 Given 2 A 5 C 7x 2 6 E 7 G A B 3 angles have equal measures 4 B 5 x 4

14 Given m4PMN mARBC Prove mAABR m4PMN m4ABC B A C

Determine the range of the following graph 12 11 10 9 8 7 6 5 4 3 2 12 11 10 9 8 7 6 2345 6 10 11 1 2 3 4 5 6 7 8 9 10 11 12

Compute 1 2 3 4 5 6 The indicates a transpose Compute 1 2 3 4 6 0 2 5 3 2 5 1 1 2 1 where semicolons notation separate rows within a matrix

1 5 1 2 x y oldu una A 7 1 2 g re x y toplam ka t r B 1 12 D 7 1 2 4 4 C 4 E 47

A HW A football is thrown by a quarterback to a receiver The points in the figure show the height of the football in feet above the ground in terms of its distance in yards from the quarterback 12 Find the coordinates of point A Then interpret the coordinates 10 in terms of the information given 14 8 What are the coordinates of point A Ay 6 4 Type an ordered pair Part 1 of 2 2 0 0 Quarterback Receiver 20 40 60 X 80 G

Evaluate 4x 5 y 1 when x is 2 less than the quotient of y and 4 and y 16 4x 5 y 1 Simplify your answer

Plot the point 0 6 in a rectangular coordinate system Plot the point 0 6 on the graph to the right E 12 Ay

Evaluate 2 L W for L 7 and W 5

Plot 4 3 on the coordinate axes Plot the point 4 3 on the graph to the right Next question 10 8 B 4 10 e 2 Ay

Use the order of operations to simplify the expression 6 8 2 6 8 2 Simplify your answer

Solve the following formula for m B mn4 m Simplify your answer

Give the ordered pair that corresponds to the point shown in the figure What are the coordinates of the point Type an ordered pair Z 6 5 4 3 2 1 6 5 4 3 2 2 3 4 6

Evaluate the expression for x 5 and y 3 5x 3y 5x 3y Simplify your answer

Solve V LWH for W W

Use an identity to find the value of the expression Do not use a calculator sec 359 tan 359 sec 359 tan 359 Type an exact answer Simplify your answer

Solve the following equation and check the solution 30 y 1 6 6 y Simplify your answer y

The graph of a function is shown below Use the graph of the function to find its average rate of change from x 1 to x 7 Simplify your answer as much as possible Nit

Write the expression in power form ax for numbers a and 6 V8x5 2 216 5 8x

Express y x 2 in polar coordinates O r sec 2 0 Or sec 2 0 tan 0 r sec 0 tan 0 O r tan e

Determine the domain of the following graph 249 900 12 Y 11 10 8 7654325 1 12 11 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 c c i i n cho n 3 4 5 6 7 9 10 11

Graph the piecewise defined function f x 3 x if x 1 4 2x if x 1 Choose the correct graph below OA Ay OB BE 57 O C 4 o O 57 O D

Use the graph of y f x to find each function value a f 3 b f 0 c f 2 d f 6 a f 3 b f 0 c f 2 d f 6 www 10 86 10 B 2 co POP

The displayed chart and accompanying histogram show the numbers of home runs hit by all baseball players in a league in the 2021 season As you can see many players hit fewer than 5 home runs while the best players hit more than 30 Complete parts a through e Click the icon to view a histogram of the home runs a Which would be the better summary of the number of home runs hit the mean or the median Why The is a better summary because the data The data are skewed to the left are unimodal and symmetric are skewed to the right Count 643 Mean 11 2 Median StdDev IQR Q1 Q3 9 10 6 13 4 17

function f x x 7 2 5 by using translations vertical or horizontal shifts Compare the graph to the graph of f x x The graph of f x x 7 2 5 is exactly the same as the graph of f x x except the graph of f x x 7 2 5 is shifted 7 units to the Choose the correct graph below OA OB O C G OD and 5 units 5

Matter is in a liquid state when its temperature is between its melting point and its boiling point Suppose that some substance has a melting point of 46 72 C and a boiling point of 328 93 C What the range of temperatures in degrees Fahrenheit for which this substance is not in a liquid state Hint C F 32 Let x represent the temperature in degrees Fahrenheit The range of temperatures for which this substance is not in a liquid state is Type an inequality or a compound inequality Simplify your answer Use integers or decimals for any numbers in the expression Round to three decimal places as needed

Your test scores in one class are 81 and 87 What possible scores can you earn on your next test to have a test average between 84 and 89 inclusive Your score on the third test must be between and inclusive Use ascending order