The graph of a piecewise-defined function is shown below.

On the interval (x≤0, x≤1, x<1, x<0) the function is given by the rule (x= - 1, y= - 1, y=1, x=1), and on the interval (x≥0, x>1, x≥1, x>0) the function is given by the rule (y= - 1, x= - 1, y= 1, x= 1).

the correct answer is =

The slope of the line is 3.5

This means that every pound

of dried fruit cost $3.50

   Slope of the line is 3.5. This means that every of dried fruit costs $3.5 per lb.

   From the graph attached,

Points shown on the graph are (4, 14) and (8, 28).

Since, slope of a line passing two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by,

Slope = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

Therefore, slope of the line passing through (4, 14) and (8, 28) will be,

Slope = [tex]\frac{28-14}{8-4}[/tex]

          = [tex]\frac{14}{4}[/tex]

          = 3.5

This means that every of dried fruit costs $3.5 per lb.

       Therefore, slope of the line is 3.5. This means that every of dried fruit costs $3.5 per lb.

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B. 2-x = 4x+3

x 2-x 4x+3

-3 5 -9
-2 4 -5
-1 3 -1
0 2 3
1 1 7
2 0 11
3 -1 15

The table shows that none of the integers from [-3,3] work because in no case does
2-x = 4x+3

To find the solution we need to rearrange the equation to the form x=n
2-x = 4x+3
2 -x + x = 4x + x +3
2 = 5x + 3
2-3 = 5x +3-3
5x = -1
x = -1/5

The only point that satisfies both equations is where x = -1/5
Find y: y = 2-x = 2 - (-1/5) = 2 + 1/5 = 10/5 + 1/5 = 11/5
Verify we get the same in the other equation
y = 4x + 3 = 4(-1/5) + 3 = -4/5 + 15/5 = 11/5

Thus the only actual solution, being the point where the lines cross, is the point (-1/5, 11/5)

C. To solve graphically 2-x=4x+3
we would graph both lines... y = 2-x and y = 4x+3
The point on the graph where the lines cross is the solution to the system of equations ...
[It should be, as shown above, the point (-1/5, 11/5)]

To graph y = 2-x make a table....
We have already done this in part B

x 2-x x 4x+3
_ __
-1 3 -1 -1
0 2 0 3
1 1 1 7

Just graph the points on a Cartesian coordinate system and draw the two lines. The solution is, as stated, the point where the two lines cross on the graph.

I hope I helped! 

I will still trying to see if I can solve them another way that might be clearer.

Answer:

Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

Counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Step-by-step explanation:

Given  : rectangle has vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4)

To find :  transformed according to the rule 90º , what is the location of q?

Solution : we have given that

vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4).

By the rule of 90º rotation clock wise rule : (x ,y ) →→ ( y , -x )

90º rotation counter clock wise rule : (x ,y ) →→ ( -y , x ).

Then   Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Therefore, Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Answer:

A

Step-by-step explanation:

hope it helps!!!!!!!

Answer:

A on edge 2020

Step-by-step explanation:

Final answer:

After converting 7/20 to a decimal, we find that Rose bought 0.35 kilogram of ginger candy, which is less than the 0.4 kilogram of cinnamon candy; therefore, Rose bought more cinnamon candy.

Explanation:

Rose bought 7/20 kilogram of ginger candy and 0.4 kilogram of cinnamon candy. To determine which she bought more of, we need to compare these amounts. The fraction 7/20 can be converted into a decimal to make the comparison easier. To convert a fraction to a decimal, you divide the numerator by the denominator.

So, 7 ÷ 20 = 0.35. This means that Rose bought 0.35 kilogram of ginger candy, which is less than the 0.4 kilogram of cinnamon candy. Therefore, Rose bought more cinnamon candy than ginger candy.

Answer:

C

Step-by-step explanation:

Cause i'm a genius

The Pythagorean theorem is applied to a right triangle with the longer leg being one inch longer than the shorter leg and the hypotenuse being nine inches longer than the shorter leg. The side lengths of the triangle are found to be 5 inches, 6 inches, and 14 inches.

The subject of this question is mathematics, specifically the part of geometry that deals with right triangles and the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (c), is equal to the sum of the squares of the other two sides (a and b), i.e., a² + b² = c².

In this problem, the longer leg (a) is represented as b = a + 1 and the hypotenuse (c) as c = a + 9. If you substitute these two equations into the Pythagorean theorem, you get (a + 1)² + a² = (a + 9)². Solving this equation gives a = 5 inches.

Substituting a = 5 inches into the expressions for b and c, we get b = 6 inches and c = 14 inches. Therefore, the sides of the triangle are 5 inches, 6 inches, and 14 inches.

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x^2 - 2x -4 =0

Using the quadratic formula -b +/- √b^2 - 4(ac) / 2a

Replace the letters with the values from the equation:

2 +/- √-2^2 -4*(1*-4) / 2*1

X = 2 +/- 2√5 / 2

x = 1 +/-√5

The answer is:  x^2 - 2x -4 =0

Answer:

16 feet

Step-by-step explanation:

The relationships between displacement (position), velocity and acceleration are:

[tex]\boxed{\boxed{\begin{array}{c}\textbf{DISPLACEMENT (s)}\\\\\text{Differentiate} \downarrow\qquad\uparrow\text{Integrate}\\\\\textbf{VELOCITY (v)}\\\\\text{Differentiate}\downarrow\qquad\uparrow \text{Integrate}\\\\\textbf{ACCELERATION (a)}\end{array}}}[/tex]

To find the distance a particle travels in its first 4 seconds of travel, we first need to determine if the particle changes direction at any point during this time.

The instant(s) when the particle changes direction is when its velocity is zero. Therefore, set the velocity function v(t) to zero and solve for t:

[tex]\begin{aligned}v(t)&=0\\-t^2+4&=0\\-t^2&=-4\\t^2&=4\\t&=\pm 2 \end{aligned}[/tex]

So, the particle changes direction at t = 2 seconds in its first 4 seconds of travel. This means that to find the distance the particle travels in its first 4 seconds of travel, we need to integrate the velocity function over the two intervals [0, 2] and [2, 4] seconds to find the particle's displacement over these intervals.

The displacement of the particle in its first 2 seconds of travel is:

[tex]\begin{aligned}\displaystyle \int^2_0 (-t^2+4)\; \text{d}t&=\left[\dfrac{-t^{2+1}}{2+1}+4t\right]^2_0\\\\&=\left[-\dfrac{t^{3}}{3}+4t\right]^2_0\\\\&=\left(-\dfrac{(2)^{3}}{3}+4(2)\right)-\left(-\dfrac{(0)^{3}}{3}+4(0)\right)\\\\&=-\dfrac{8}{3}+8+0-0\\\\&=\dfrac{16}{3}\end{aligned}[/tex]

The displacement of the particle in its next 2 seconds of travel is:

[tex]\begin{aligned}\displaystyle \int^4_2 (-t^2+4)\; \text{d}t&=\left[\dfrac{-t^{2+1}}{2+1}+4t\right]^4_2\\\\&=\left[-\dfrac{t^{3}}{3}+4t\right]^4_2\\\\&=\left(-\dfrac{(4)^{3}}{3}+4(4)\right)-\left(-\dfrac{(2)^{3}}{3}+4(2)\right)\\\\&=-\dfrac{64}{3}+16+\dfrac{8}{3}-8\\\\&=-\dfrac{32}{3}\end{aligned}[/tex]

The negative value means that the particle is travelling in the opposite direction.

So, the particle travels 16/3 feet in its first 2 seconds of travel, changes direction at t = 2 seconds, and travels 32/3 feet in the opposite direction in the next 2 seconds of travel.

Therefore, the total distance the particle travelled is the sum of the absolute values of the two displacements:

[tex]\textsf{Distance}=\dfrac{16}{3}+\dfrac{32}{3}=\dfrac{48}{3}=16\;\sf feet[/tex]

So, the particle travels 16 feet in its first 4 seconds of travel.

The distance a particle travels in the first 4 seconds, moving according to the velocity equation v(t)= -t² + 4, is found by integrating the velocity function over that interval, which yields 5.33 feet.

To find the distance a particle travels given its velocity equation v(t)= -t² + 4 (in feet/sec), we need to integrate the velocity function over the desired time interval.

Since velocity is the rate of change of position with respect to time, the integral of the velocity function from 0 to 4 seconds will give us the displacement of the particle in that time interval.

The integral of v(t) from 0 to 4 seconds can be computed as follows:

However, the absolute value of -5.33 feet is needed, because distance must be positive. Hence, the particle travels 5.33 feet in the first 4 seconds.

Kalahira is correct but the letters at the end might confuse you.

Correct answer:

To calculate the cost of diesel for Steve's trip from Leeds to Edinburgh, divide the distance by the car's mpg to find the number of gallons needed. Then, multiply the gallons by the cost of diesel per gallon to get the total cost.

To determine the cost of diesel for Steve's trip, we need to calculate the number of gallons he will need and then multiply by the cost per gallon. Steve's car does 42 miles per gallon (mpg), and he is driving a distance of 215 miles. Therefore, he will need 215/42 = 5.12 gallons of diesel. Since a gallon is 4.55 liters, Steve will need 5.12 * 4.55 = 23.36 liters of diesel. The cost of diesel per liter is £1.38, so the total cost for diesel will be 23.36 * £1.38 = £32.19.

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The statement that is always true is: D. adjacent linear pairs of angles are supplementary.

Linear pairs of angles are angles that lie on the straight line. The sum of a linear pair is always 180 degrees, thus, they are regarded as supplementary angles.

Therefore, the statement that is always true is: D. adjacent linear pairs of angles are supplementary.

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The given linear equation will have only one possible solution

A function is a relation between a dependent and independent variable. We can write the examples of functions as -

y = f(x) = ax + b

y = f(x, y, z) = ax + by + cz

Given is to find the total number of solutions for the linear equation -

10(x + 6) 2 - 9 = (15x + 1) 3 - 8

A linear equation is a equation of degree one. The number of solutions of a equation are equal to its degree. So, the given linear equation will have only one solution.

Therefore, the given linear equation will have only one possible solution.

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David’s score at the end of round 3 is calculated by adding the net points gained during the round to his initial score, resulting in a final score of 2000 points.

Calculating David's Score

To determine David's score at the end of round 3, we need to account for both his correct answers and incorrect answers during the round.

David's initial score at the start of round 3 is -1500 points.

During round 3:

He answered five 1000-point questions correctly. Each correct answer adds 1000 points. So, 5 correct answers add:

5 × 1000 = 5000 points

He answered three 500-point questions incorrectly. Each incorrect answer subtracts 500 points. So, 3 incorrect answers subtract:

3 × 500 = 1500 points

Next, we calculate the net change in his score by adding the points gained and subtracting the points lost:

Net change = 5000 points (gained) - 1500 points (lost) = 3500 points

Finally, we add this net change to his initial score:

Final score = -1500 points (initial score) + 3500 points (net change) = 2000 points

Thus, David's score at the end of round 3 is 2000 points.

The lines through points P and Q and points R and S are not perpendicular.

To determine if the line through points P(-8, -10) and Q(-5, -12) is perpendicular to the line through points R(9, -6) and S(17, -5), we need to calculate the slopes of both lines and check if they are negative reciprocals of each other. The slope of a line (m) through two points (x1, y1) and (x2, y2) is given by the formula m = (y2 - y1) / (x2 - x1).

For line PQ:

mPQ = (-12 + 10) / (-5 + 8) = -2 / 3

For line RS:

mRS = (-5 + 6) / (17 - 9) = 1 / 8

Now, the product of the slopes of two perpendicular lines is -1. Let's check if the product of mPQ and mRS is -1:

mPQ * mRS = (-2 / 3) * (1 / 8) = -2 / 24 = -1 / 12

Since -1 / 12 is not equal to -1, the lines are not perpendicular. Therefore, the line through points P and Q is not perpendicular to the line through points R and S.

Using a zero as a placeholder so that both decimals ahve the same number of digits after their decimal point is a good strategy while adding decimals because it helps you line the decimals up in the correct way to get the correct sum.