Pls answer this question <br>don't answer with a link

Write a piecewise function for the graph

netineya [11]

Answer:

You can used it to help me out with a good time with this and

Step-by-step explanation:

5 0

2 years ago

What are the solution(s) to the quadratic equation 50 - x² = 0?

grandymaker [24]

Answer:

The answer is C.

Step-by-step explanation:

$50-x^2=0$

$x^2=50$

$x=\pm \sqrt{50} =\pm \sqrt{25*2}=\pm 5\sqrt{2}$

The answer is C (I am assuming that it isn't 5/2).

7 0

2 years ago

Show that the line 4y = 5x-10 is perpendicular to the line 5y + 4x = 35

Shkiper50 [21]

Step-by-step explanation:

<h2><em><u>concept :</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10or, y = (5/4)x(5/2).</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>1</em><em>)</em></h2><h2 /><h2><em><u>5y + 4x = 35</u></em></h2><h2 /><h2><em><u>5y + 4x = 35ory = (-4/5)x + 7.</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>2</em><em>)</em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1Hence, the lines are perpendicular.</u></em></h2>

8 0

2 years ago

Solve the system of linear equations below.

Reika [66]

Answer:

x = 1 , y = 3 thus: A is your Anser

Step-by-step explanation:

Solve the following system:

{2 x + y = 5 | (equation 1)

x + y = 4 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y = 5 | (equation 1)

0 x+y/2 = 3/2 | (equation 2)

Multiply equation 2 by 2:

{2 x + y = 5 | (equation 1)

0 x+y = 3 | (equation 2)

Subtract equation 2 from equation 1:

{2 x+0 y = 2 | (equation 1)

0 x+y = 3 | (equation 2)

Divide equation 1 by 2:

{x+0 y = 1 | (equation 1)

0 x+y = 3 | (equation 2)

Collect results:

Answer: {x = 1 , y = 3

8 0

2 years ago

Read 2 more answers

W is the midpoint of VX and Z is the midpoint of VY . If XY=p and WZ=p–30, what is WZ?

juin [17]

$\large{ \tt{❃ \: EXPLANATION}} :$

You'll have to know : Mid-point theorem states that A straight line segment joining the mid-points of any two triangle is parallel to the third side and it is equal to half of the length of the third side.