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If f(x)=-4x^2-6x-1 and g(x)=-x^2-5x+3, fine (f-g)(x)

NeX [460]

Answer:

$\large\boxed{(f-g)(x)=-3x^2-x-4}$

Step-by-step explanation:

$(f-g)(x)=f(x)-g(x)$

$f(x)=-4x^2-6x-1,\ g(x)=-x^2-5x+3\\\\(f-g)(x)=(-4x^2-6x-1)-(-x^2-5x+3)\\\\(f-g)(x)=-4x^2-6x-1-(-x^2)-(-5x)-3\\\\(f-g)(x)=-4x^2-6x-1+x^2+5x-3\qquad\text{combine like terms}\\\\(f-g)(x)=(-4x^2+x^2)+(-6x+5x)+(-1-3)\\\\(f-g)(x)=-3x^2-x-4$

5 0

4 years ago

When finding 433 - 249 using place value. What do you subtract first?

Ksenya-84 [330]

You subtract the ones place, then the tens, and lastly the hundreds place. Your answer will be 184 if you are looking for the answer as well.

5 0

3 years ago

Tamara and Amir shared a candy bar. Tamara ate two fifths. Amir ate two fifths. How much is left?

marin [14]

Answer:

1/5

Step-by-step explanation:

Both Amir and Tamara ate 2/5 each.

2/5 + 2/5 = 4/5

If 4/5 of the candy bar was eaten, 1/5 of the candy bar remains because

5/5 - 4/5 = 1/5

(Remember that 5/5 just equals one whole, representing the whole candy bar)

7 0

3 years ago

Graph y ≤ (x + 2)^2.

Leokris [45]

Answer:

Step-by-step explanation:

8 0

3 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

3 years ago