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3 years ago

Someone please help me with this ASAP

Mathematics

2 answers:

makkiz [27]3 years ago

6 0

Answer:

y=-4x-1

Step-by-step explanation:

y=mx+b

m is the slope

b is the y-intercept

LenaWriter [7]3 years ago

4 0

Answer: y = -4x -1

Step-by-step explanation:

Note that the equation for slope-intercept is y = mx + b

m = slope

b = y-intercept

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Ok last question then im done

kkurt [141]

55.. i am pretty sure. if i’m wrong you could slap me

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Dan tan 31 miles more than Jennifer last week. Dan ran 138 miles. How many miles did Jennifer run?

PtichkaEL [24]

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What is the sum of all values of m that satisfy 2m (squared) -16m+8=0?

vodomira [7]

<h2>Steps:</h2>

So for this, we will be completing the square to solve for m. Firstly, subtract 8 on both sides:

$2m^2-16m=-8$

Next, divide both sides by 2:

$m^2-8m=-4$

Next, we want to make the left side of the equation a perfect square. To find the constant of this perfect square, divide the m coefficient by 2, then square the quotient. In this case:

-8 ÷ 2 = -4, (-4)² = 16

Add 16 to both sides of the equation:

$m^2-8m+16=12$

Next, factor the left side:

$(m-4)^2=12$

Next, square root both sides of the equation:

$m-4=\pm \sqrt{12}$

Next, add 4 to both sides of the equation:

$m=4\pm \sqrt{12}$

Now, while this is your answer, you can further simplify the radical using the product rule of radicals:

Product rule of radicals: √ab = √a × √b

√12 = √4 × √3 = 2√3.

$m=4\pm 2\sqrt{3}$

<h2>Answer:</h2>

In exact form, your answer is $m=4\pm \sqrt{12}\ \textsf{OR}\ m=4\pm 2\sqrt{3}$

In approximate form, your answers are (rounded to the hundreths) $m=7.46, 0.54$

6 0

3 years ago

At which points are the tangents drawn to the ellipse x 2 + y 2 = [ a ] x + [ a ] y parallel to

In-s [12.5K]

The given equation of the ellipse is x^2 + y^2 = 2 x + 2 y

At tangent line, the point is horizontal with the x-axis therefore slope = dy / dx = 0

<span>So we have to take the 1st derivative of the equation then equate dy / dx to zero.</span>

x^2 + y^2 = 2 x + 2 y

x^2 – 2 x = 2 y – y^2

(2x – 2) dx = (2 – 2y) dy

(2x – 2) / (2 – 2y) = 0

2x – 2 = 0

x = 1

To find for y, we go back to the original equation then substitute the value of x.

x^2 + y^2 = 2 x + 2 y

1^2 + y^2 = 2 * 1 + 2 y

y^2 – 2y + 1 – 2 = 0

y^2 – 2y – 1 = 0

Finding the roots using the quadratic formula:

y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1

y = 1 ± 2.828

y = -1.828 , 3.828

<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828) and (1, 3.828).</span>

3 0

3 years ago