Which set of parametric equations represents the function y=x^2+4x-5? Select all that apply.

Which Set Of Parametric Equations Represents The Function Y=x^2+4x-5? Select All That Apply.

Answers

Answer 1

Solution

- The way to solve the equation is to take the expression for x i.e. x = 2t, and substitute into the expression for y(x).

- The result must be the corresponding y-value in terms of t.

- This is done below:

Option A:

[tex]\begin{gathered} x=2t \\ y(x)=x^2+4x-5 \\ \\ \text{ put }x(t)=2t \\ \\ y(x(t))=(2t)^2+4(2t)-5 \\ y(x(t))=y(t)=4t^2+8t-5 \\ \\ \therefore y(t)=4t^2+8t-5\text{ \lparen OPTION A\rparen} \end{gathered}[/tex]

Option B:

[tex]\begin{gathered} x=t+1 \\ y=x^2+4x-5 \\ \\ y(x(t))=y(t)=(t+1)^2+4(t+1)-5 \\ t^2+2t+1+4t+4-5 \\ y(t)=t^2+6t\text{ \lparen NOT IN THE OPTIONS\rparen} \end{gathered}[/tex]

Option C:

[tex]\begin{gathered} x=t-3 \\ y=x^2+4x-5 \\ \\ y(x(t))=(t-3)^2+4(t-3)-5 \\ =t^2-6t+9+4t-12-5 \\ =t^2-2t-8\text{ \lparen NOT IN THE OPTIONS\rparen} \end{gathered}[/tex]

Option D:

[tex]\begin{gathered} x=t^2 \\ y=x^2+4x-5 \\ \\ y(x(t))=(t^2)^2+4(t^2)-5 \\ =t^4+4t^2-5\text{ \lparen NOT IN THE OPTIONS\rparen} \end{gathered}[/tex]

Option E:

[tex]\begin{gathered} x=t+1 \\ y=x^2+4x-5 \\ \\ y(x(t))=(t+1)^2+4(t+1)-5 \\ =t^2+2t+1+4t+4-5 \\ =t^2+6t\text{ \lparen OPTION E IS CORRECT\rparen} \end{gathered}[/tex]

Final Answer

The answers are OPTIONS A AND E


Related Questions

Joe goes running in the park. He runs 3 miles and does it in 42 minutes. How many minutes doe it take him to run a mile? This topic is distance = rate x time

Answers

You must use this formula:

[tex]d=rt[/tex]

Where "d" is the distance, "r" is the rate and "t" is time.

If you solve for "r":

[tex]r=\frac{d}{t}[/tex]

If you solve for "t":

[tex]t=\frac{d}{r}[/tex]

Knowing that Joe runs 3 mile in 42 minutes, you can find "r". Notice that:

[tex]\begin{gathered} d=3mi \\ t=42\min \end{gathered}[/tex]

Then:

[tex]r=\frac{3mi}{42\min}=0.0714\frac{mi}{\min}[/tex]

Knowing the rate, you can set up the following in order to find the time in minutes it takes Joe to run a mile:

[tex]\begin{gathered} d=1mi \\ r=0.071\frac{mi}{\min} \end{gathered}[/tex]

Substituting values into the formula for calculate the time, you get:

[tex]t=\frac{1\min}{0.0714\frac{mi}{\min}}=14\min [/tex]

The answer is: It takes him 14 minutes to run a mile.

Find the distance between the points. Round to the nearest tenth if necessary. (3, 7), (-5, -7) Distance?

Answers

To find the distance between both points you have to apply pythagoras theorem.

First draw both points and form a rigth triangle with the distance between them as the hypothenuse:

The length of the base of the triangle "a" is determined by the difference between the x-coordinates of both points:

[tex]a=x_2-x_1=3-(-5)=3+5=8[/tex]

The heigth of the triangle "b" is determined by the difference between the y-coordinates of both points:

[tex]b=y_2-y_1=7-(-7)=7+7=14[/tex]

Now using phytagoras theorem you can calculate the length of the hypotenuse as:

[tex]\begin{gathered} a^2+b^2=c^2 \\ (8)^2+(14)^2=c^2 \\ c^2=260 \\ c=\sqrt[]{260} \\ c=2\sqrt[]{65}=16.12 \end{gathered}[/tex]

The distance between points (3,7) and (-5,-7) is 2√65

A length measure can never be more than one half unit in error. why is this the case?can someone please answer this question.

Answers

Answer:

This is because the degree of accuracy is half a unit each side of the unit of measure

[tex]\text{When an instrument measures in '1' s any value betwe}en\text{ 6}\frac{1}{2}\text{ and 7}\frac{1}{2\text{ }}\text{ is measured as 7}[/tex]

Find the area of the shaded region assume all angles are right angles

Answers

The given figure is of a rectangle which is enclosed in the large rectangle.

Area of rectangle = Length x Width

Dimension of large rectangle, 10 and 20.

Area of larger rectangle = 10 x 20

Area of larger rectangle = 200

Dimension of the small rectangle, 14 and 6.

Area of small rectangle = 14 x 6

Area of small rectangle = 84

Area of shaded region = Area of large rectangle - Area of small rectangle

Area of shaded region = 200 - 84

Area of shaded region = 116

Area of shaded region is 116 unit²

Hi , can you help me to solve this problem please.

Answers

Answer:

The polynomials are classified as shown in the image below

Hello. I am trying to help my 9th grade daughter with text corrections. It has been over 20 yrs since I had Algebra 1 and Im a bit rusty. She gets easily frustrated especially in math so Im trying to do some of the leg work before going over how to do it with her. I appreciate your help in advance.

Answers

The half-life of a radioactive substance is given 3 hours.

Given the initial amount of substance is 800 grams. After 3 hours, the substance becomes half that is 400 grams. Then again after 3 more hours, the substance becomes half again that is 200 grams. Again after three hours, the substance becomes half that is 100 grams.

Thus, the amount of radioactive material after 9 hours is 100 grams.

Your team has carefully researched and selected two possible painting companies. Pro Painters charge $200 per hour plus $6000 in material fees. Illusion Ltd charges $150 per hour plus $8000 in material fees.Create a graph of the cost for both companies using the grid below. Circle the point of intersection. Be sure your lines are properly identified.

Answers

Given:

• Pro Painters:

Charge per hour = $200

Material fees = $6000

• Illusion Ltd:

Charge per hour = $150

Material fees = $8000

Let's create a graph of the cost for both companies.

Represent each situation using the slope-intercept form:

y = mx + b

In this case, y represents the total charge, m is the charge per hour, x represents the number of hours, and b represents the material fees.

We have the following:

• Equation for Pro Painters:

y = 200x + 6000

• Equation for Illsion Ltd:

y = 150x + 8000

To graph let's create two points on each equation.

We have:

• Pro painters:

y = 200x + 6000

When x = 1: y = 200(10) + 6000 = 8000

When x = 3: y = 200(30) + 6000 = 12000

We have the points:

(x, y) ==> (10, 8000), (30, 12000)

Plot the points and connect them using a straight line.

• Illusion Ltd:

y = 150x + 8000

When x = 2: y = 150(20) + 8000 = 11000

When x = 4: y = 150(40) + 8000 = 14000

We have the points:

(x, y) ==> (20, 11000), (40, 14000)

Plot the points and connect them using a straight line.

We have the graph below:

The green line represents the cost for Pro Painters

The blue line represents the cost for Illusion Ltd.

From the graph, the point of intersection is (40, 14000).

This means at 40 hours, the cost for both companies will be the same ($14,000)

ANSWER:

• Equation for Pro painters: , y = 200x + 6000

,

• Equation for Illusion Ltd: , y = 150x + 8000

,

• Point of intersection: (40, 14000)

Which expressions are equivalent to the one below? Check all that apply.log3 3+ log3 27A. log3 81B. log3 (3^4)C. 4D. log 10

Answers

The given expression is

[tex]log_33+log_327[/tex]

We will use the rule

[tex]log_ba+log_bc=log_b(ac)[/tex][tex]\begin{gathered} log_33+log_327=log_3(3\times27) \\ \\ log_3(3\times27)=log_3(81) \end{gathered}[/tex]

Since 81 = 3 x 3 x 3 x 3, then

[tex]\begin{gathered} 81=3^4 \\ log_3(81)=log_3(3^4) \end{gathered}[/tex]

We will use the rule

[tex]log_b(a^n)=nlog_b(a)[/tex][tex]undefined[/tex]

Consider the function f(x) = 5 - 4x ^ 2, - 5 <= x <= 1 .

Answers

Given: A function-

[tex]f(x)=5-4x^2,\text{ }-5\leq x\leq1[/tex]

Required: To determine the absolute maxima and minima of the function.

Explanation: The given function is-

[tex]f(x)=5-4x^2[/tex]

Differentiating the function,

[tex]f^{\prime}(x)=-8x[/tex]

Setting f'(x)=0 gives-

[tex]\begin{gathered} -8x=0 \\ \Rightarrow x=0 \end{gathered}[/tex]

So we have to check the function at the boundary points of the interval [-5,1] and x=0 as follows-

Hence, the absolute maximum is 5 at x=o, and the minimum is -95 at x=-5.

Final Answer: The absolute maximum value is 5, and this occurs at x=0.

The absolute minimum value is -95, and this occurs at x=-5.

For the data set 1,7,7,7,8, the mean is 6. What is the mean absolutedeviation?O A. The mean absolute deviation is 10.O B. The mean absolute deviation is 6.O c. The mean absolute deviation is 2.O D. The mean absolute deviation is 5.

Answers

The mean absolute deviation is given by:

[tex]\frac{\sum ^{}_{}\lvert x_i-\bar{x}\rvert}{n}[/tex]

where xi represent each data, x bar the mean and n the number of data we have. Then:

[tex]\begin{gathered} \frac{\lvert1-6\rvert+\lvert7-6\rvert+\lvert7-6\rvert+\lvert7-6\rvert+\lvert8-6\rvert}{5} \\ =\frac{\lvert-5\rvert+\lvert1\rvert+\lvert1\rvert+\lvert1\rvert+\lvert2\rvert}{5} \\ =\frac{5+1+1+1+2}{5} \\ =\frac{10}{5} \\ =2 \end{gathered}[/tex]

Therefore the mean absolute value is 2 and the answer is C.

4. Each month for 2 months, Kyle buys a pack of 8 replacement tires for his remote-control car. At the end of each month, he has 1 tire left. Explain how the numerical expression for the number of tires Kyle uses in 2 months compares to the numerical expression for the number of tires he uses in 1 month.

Answers

Each month Kyler buys a pack of 8 replacement tires.

He does this for 2 months.

At the end of each month he has 1 tire left.

So the numerical expression for the number of tires he uses per month will be;

Number of tires used in 1st month = 8-1 =7

Number of tires used in the 2nd month = 8 - 1= 7

Total number of tires used in two months = 7*2 = 14

Tires left in two months = 1+ 1 = 2

Comparison : The number of tires used in two months is twice that used in one month.

Write 7.916 x 10-7 in decimal form.

Answers

In order to convert the number to decimal form, we need to look at the exponent of the number 10 multiplying the number 7.916.

The exponent is equal to -7, which means we will need to add 7 times the number 0 to the left of the number 7.916. Also, the decimal point will be moved 7 positions to the left.

So we have:

[tex]7.916\cdot10^{-7}=0.0000007916[/tex]

Can you explain.Use the intermediate value theorem for polynomials to show that the polynomial function has a real zero between the numbers given.f(x) = -6x^4+5x^2+4;-2 and -1

Answers

SOLUTION:

We are to show that the given polynomial function has a real zero between the numbers given.

[tex]f(x)=-6x^4+5x^2\text{ + 4}[/tex]

At x = -2, we substitute -2 for x in the given function;

[tex]\begin{gathered} f(-2)=-6(-2)^4+5(-2)^2\text{ + 4} \\ f(-2)\text{ = -6(16) + 5(4) + 4} \\ f(-2_{})\text{ = -96 + 20 + 4} \\ f(-2)\text{ = -72} \end{gathered}[/tex]

At x = -1, we substitute -1 for x in the given function;

[tex]\begin{gathered} f(-1)=-6(-1)^4+5(-1)^2\text{ + 4} \\ f(-1)\text{ = -6(1) + 5(1) + 4} \\ f(-1)\text{ = -6 + 5 + 4} \\ f(-1)\text{ = 3} \end{gathered}[/tex]

CONCLUSION:

Since the function f went from -72 to +3 over the interval of -2 to -1, that means it must have passed through zero.

What is the value of x in the solution to the system of equations below?2x+3y=112x+y=1

Answers

Answer:

x=-2

Explanation:

Given the system of equations:

[tex]\begin{gathered} 2x+y=1 \\ 2x+3y=11 \end{gathered}[/tex]

We use the method of elimination by subtracting.

This gives us:

[tex]\begin{gathered} -2y=-10 \\ y=\frac{-10}{-2} \\ y=5 \end{gathered}[/tex]

We then substitute y=5 into any of the equations to solve for x.

[tex]\begin{gathered} 2x+y=1 \\ 2x+5=1 \\ 2x=1-5 \\ 2x=-4 \\ x=-\frac{4}{2} \\ x=-2 \end{gathered}[/tex]

Therefore, the value of x in the solution to the system of equations is -2.

A sofa and a love seat together costs $600. The sofa costs $75 less than double the love seat. How much do they each cost The equation

Answers

To solve this problem we need to create an equation, where the unkown variable, x, represents the cost for the love seat. We know that the sofa costs $75 less than the love seat, therefore we have:

[tex]y=x-75[/tex]

The cost for both pieces of furniture together is equal to $600. So if we add them we have:

[tex]x+y=600[/tex]

We can swap the expression for y on the second equation.

[tex]\begin{gathered} x+(x-75)=600 \\ x+x-75=600 \\ 2x-75=600 \\ 2x=675 \\ x=\frac{675}{2}=337.5 \end{gathered}[/tex]

Now we know that the love seat costs $337.5. We will use the first equation to find the cost of the sofa.

[tex]y=337.5-75=262.5[/tex]

The sofa costs $262.5.

could you please help me out with a question

Answers

At figure , Diameter = √ 24^2 + 7^2 = √ 625 = 25

Then ANSWERS ARE

Circumference = π• D = 3.14 x 25 = 78.5 cm

Area of circle = π• D^2/4 = 3.14 x 25^2/4 = 490.6 cm2

Simplify the expression.

9. (x^-3)^-5x^6

Answers

Answer: x to the power of 21

Step-by-step explanation:

the area of a trapezium is 1680 sq cm. One of the parallel sides is 64 cm and the perpendicular distance between the parallel sides is 28 cm. find the length of the other parallel side.​

Answers

Answer:

The missing side length is 56

Step-by-step explanation:

1680 = 1/2 · (64 + x) · 28

1680 · 2 = 28 · (64 + x)

3360 = 1792 + 28x

28x = 3360 - 1792

28x = 1568

x = 1568 ÷ 28

x = 56

Hope this helps.

Factor 4a²x - 4ax - 8x.

Answers

Answer:

4x(a+1)(a-2)

Explanation:

Given the polynomial:

[tex]4a^2x-4ax-8x[/tex]

First, factor out 4x in all the terms:

[tex]=4x(a^2-a-2)[/tex]

Next, factor the expression in the parenthesis:

[tex]\begin{gathered} =4x(a^2-a-2) \\ =4x(a^2-2a+a-2) \\ =4x[a(a-2)+1(a-2)] \\ =4x(a+1)(a-2) \end{gathered}[/tex]

The factored form of the polynomial is 4x(a+1)(a-2).

Refer to the figure below to answer the following questions: (a) When placed in Quadrant ), name the coordinates of point T that forms parallelogram QTRS. (b) When placed in Quadrant II, name the coordinates of point T that forms parallelogram QRST. (c) When placed in Quadrant IV, name the coordinates of point T. that forms parallelogram QRTS. Given Points Q(-1,3), R(3.0), and S(-2,-1) Q T. S

Answers

A parallelogram is a quadilateral that has two pairs of parallel sides. The opposite sides of a parallelogram are equal.

Given the points:

Q(-1,3), R(3,0), and S(-2,-1)

a) When placed in quadrant I, let's find the point T that forms a parallellogram.

Here the distance QS and RT must be equal.

Use the distance formula:

[tex]d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}[/tex]

The point of T that forms a parallellogram when placed in quadrant I is:

T(4, 4)

From point R

b) When placed in Quadrant II, let's find the point T that forms a parallellogram.

We have:

T(-6, 2)

From point Q, make a movement 5 units left and 1 unit down

The point of T that forms a parallellogram when placed in quadrant II is:

T(-6, 2)

c) When placed in quadrant IV, let's find the point T that forms a parallelogram.

We have:

T(2, -4)

From point R, make a movement of down 4 units and left 1 unit.

The point of T, that forms a parallelogram when placed in quadrant IV is:

T(2, -4)

ANSWER:

a) (4, 4)

b) (-6, 2)

c) (2, -4)

Alex and George are each charged a constant rate for every text they send on their cell phones, shown in the table below:Number ofTexts (0)AmountCharged (a)$0.20$0.40$0.6023At the end of the month, Alex learned that he had sent 150 texts and paid $30 for his bill. George learned that he had sent 125 texts and paid $6.25.Who paid the correct amount? How do you know?

Answers

the payment rate is $0.2 per texts. So Alex should pay

[tex]150\cdot0.2=30[/tex]

so the paid is correct for Alex. And we know because it follows the payment rate

Sketch a diagram of a 315° angle in standard position and indicate the measure of its reference angle.

Answers

Explanation:

The angle in the question is given below as

[tex]315^0[/tex]

The angle is found the QUADRANT IV

Hence,

The sketch of the angle in standard positing will be given in the image below

[tex]360-315=45^0[/tex]

Therefore,

The reference angle is

[tex]45^0[/tex]

Hi I need help with this math problem, i’m in high school calculus 1

Answers

Step 1:

When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression.

Step 2

Parent function y = f(x)

In general, a vertical stretch is given by the equation

y=bf(x). If b>1, the graph stretches with respect to the y-axis, or vertically. If b<1, the graph shrinks with respect to the y-axis.

The function becomes y = 1.4f(x) when trainsform vertically

The function is shifted 3 units to the left and it becomes y = 1.4f(x + 3)

Final answer

y = 1.4f(x + 3)

Solve for h: A = (1/2)*b*h*O h = 2*A*bO h = A *(b/2)O h = (2*A)/b0 h = (2+b)/A

Answers

[tex]\begin{gathered} A=\frac{1}{2}bh \\ \text{cross multiply} \\ 2A=bh \\ \frac{2A}{b}=h \\ \\ h=\frac{2A}{b} \end{gathered}[/tex]

This table represents the relationship between x and y described by the equation.y=-x1012141618SY6789Which list represents the dependent values in the table?5,6,7,8,95, 6, 7, 8, 9, 10, 12, 14, 16, 1810, 12, 14, 16, 181,2,3,4,5

Answers

ANSWER :

A. 5, 6, 7, 8, 9

EXPLANATION :

From the problem, we have the function :

[tex]y=\frac{1}{2}x[/tex]

y is the dependent variable and

x is the independent variable.

So the dependent values are the y values.

That will be 5, 6, 7, 8, 9

Write an equation for the line that contains (-81, 17) and is perpendicularto the graph 9(2x - 4) - 6(2y - 3) = 4y +2Help please! Big test tomorrow

Answers

To write the equation of a paerpendicular line that cross a given point we first need the slope of the given line, then we transform into the spole of the perpendicular line and find the intercept using the given point.

So, we want an equation like this:

[tex]y=ax+b[/tex]

And we need "a" and "b". First, let's rewrite the given equation in the slope-interscept form:

[tex]\begin{gathered} 9(2x-4)-6(2y-3)=4y+2 \\ 18x-36-12y+18=4y+2 \\ 18x-12y-18=4y+2 \\ 18x-18-2=4y+12y \\ 18x-20=16y \\ y=\frac{18}{16}x-\frac{20}{16} \\ y=\frac{9}{8}x-\frac{5}{4} \end{gathered}[/tex]

This is equivalent to the given graph. 9/8 is the slope. To get the slope of the perpendicular line, we invert it and change its sign. So "a" (the slope of the perpendicular line) is:

[tex]a=-\frac{1}{\frac{9}{8}}=-\frac{8}{9}[/tex]

Now we got:

[tex]y=-\frac{8}{9}x+b[/tex]

To find "b", we input the values of the point we want it to contain, which is (-81,17):

[tex]17=-\frac{8}{9}(-81)+b[/tex]

And we solve for b:

[tex]\begin{gathered} 17=\frac{8\cdot81}{9}+b \\ 17=8\cdot9+b \\ b=17-8\cdot9 \\ b=17-72 \\ b=-55 \end{gathered}[/tex]

So, the equation is:

[tex]y=-\frac{8}{9}x-55[/tex]

six fifths, eight ninths, 0.5, forty percent?

Answers

Answer:

I'm assuming this is a greatest to least, but in case it was not, I put least to greatest, too.

Step-by-step explanation:

Greatest to least:

6/5, 8/9, 0.5, 40%

Least to greatest:

40%, 0.5, 8/9, 6/5

Hope this helps!

Model With Mathematics An archer shoots
an arrow to a height (meters) given by the
equation y = -5t2 + 18t - 0.25, where t is
the time in seconds. A target sits on a hill
represented by the equation y = 0.75x - 1.
At what height will the arrow strike the
target, and how long will it take?

Answers

The arrow strike the target within time of  = 3.49s on height  of = 1.62 m

What is quadratic equation?A quadratic equation is a quadratic algebraic expression of the form ax2 + bx + c = 0. The word quadratic comes from the word quad, which means square. There are many scenarios where quadratic equations are used. Did you know that when a rocket launches, its trajectory is described by a quadratic equation?In addition, quadratic equations have many uses in physics, engineering, astronomy, and more. 4A quadratic equation is a quadratic equation in x for which x has at most two answers. These two answers to x are also called the roots of the quadratic equation and are denoted by (α, β). In the next content, you will learn  about roots of quadratic equations. The quadratic equation is his second algebraic equation of x. The first condition of the quadratic equation is that the coefficient of the x2 term is nonzero (a ≠ 0).Calculation

y1 = -5t² + 18t -0.25

y2 = 0.75t - 1

the arrow will strike the target when y1 = y2

-5t² + 18t -0.25 = 0.75t - 1

t = 3.49s(approx.)

the height y = 1.62m

learn more about quadratic equation here :
brainly.com/question/17177510

#SPJ1

4) Which of the following could represent the lengths of the sides of a right triangle? Hint: Remember Pythagorean Triple :a) 3,4,5b) 5,12,12c) 15,30,45d) 24,32,40

Answers

We have to find which of the following could represent the lengths of the sides of a right triangle.

To be a right triangle, the lengths a, b and c have to satisfy the Pithagorean theorem:

[tex]a^2+b^2=c^2[/tex]

Of course, c has to be the largest of the sides.

We can write for the first option:

[tex]\begin{gathered} 3^2+4^2=5^2 \\ 9+16=25 \\ 25=25 \end{gathered}[/tex]

As the expression is satisfied, we can conclude that the triangles with sides 3, 4 and 5 is a right triangle.

Option B (5,12,12) can not be a right triangle, as it has 2 largest sides. It can only have one, that is the hypothenuse. NOTE: it can have two equal smallest sides, but no two largest.

Option C is 15, 30 and 45. We test the equation:

[tex]undefined[/tex]

I need help with this work question 10Find the area of each regularpolygon. Leave your answer insimplest form.

Answers

Given:

Number of sides in octagon = 8

Length of apothem = 14.1

Side length = 11.7

Required: Area

Explanation:

The area of a regular polygon is one-half the product of its apothem and its perimeter.

Here, the area of the regular octagon is

[tex]\begin{gathered} A=\frac{1}{2}ap \\ =\frac{1}{2}\times14.1\times8\times11.7 \\ =659.88 \end{gathered}[/tex]

Final Answer: Area of the regular octagon is 659.88 square units.

Other Questions
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