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3 years ago

Write an equation for the line that is parallel to the given line and passes through the given point Y= 1/2 x -8 (-6 ,-17)

1 answer:

8 0

Parallel to : y = 1/2x - 8

Passes through (-6, -17)

Our given slope is 1/2.

Plug the given coordinates into point-slope form.

y - y₁ = m(x - x₁)

y - (-17) = 1/2(x - (-6))

Simplify.

y + 17 = 1/2(x + 6) ← Your answer.

However, sometimes teachers want this simplified even further.

So, simplify this.

y + 17 = 1/2x + 3

Subtract 17 from both sides.

y = 1/2x + 3 - 17

Simplify.

y = 1/2x - 14 ← Slope-intercept form.

Standard form → -1/2x + y = -14

~Hope this helped!~

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Factor the trinomial x^2 - 7x + 10

Valentin [98]

Answer:

(x-5) (x-2)

Step-by-step explanation:

x^2 (-2x-5x) +10

basically you will write the equation down and then to double check multiply the x and then the -5 into (x-2). this can be repetitive if you get it wrong so try to do numbers that make sense.

7 0

3 years ago

Tim install 50 square feet of his floor in 45 mins. At this rate how long does it take him to install 495 square feet. Is that 9

Nimfa-mama [501]

Answer:

the answer is 445.5

Step-by-step explanation:

what I did was divide 50 by 495 and got 9.9 then since it took 45 minutes every 50 square feet of floor i multiplied 45 x 9.9 and got 445.5

6 0

3 years ago

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scZoUnD [109]

Answer:

Step-by-step explanation:

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5 0

3 years ago

Which is a factor of 2x^2y^4-10x2y+14y3-70

VARVARA [1.3K]

2x^2y(y^3-5)+14(y^3-5)
(2x^2y)+14(y^3-5)
2(x^2y+7)(y^3-5)
x^2y+7 Is a factor

3 0

3 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

$\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}$

7 0

2 years ago