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3 years ago

9

Which is the equation of a line that is perpendicular to 8x + 6y = –5?

1 answer:

IrinaVladis [17]3 years ago

4 0

The perpendicular line will have an equation of y=3/4x-3/4

To find this, we first have to solve our equation for slope intercept form.

8x + 6y = -5 ----> subtract 8x

6y = -8x + -5 ----> divide by 6

y = -4/3x - 5/6.

So we know the slope of this equation to be -4/3. Since perpendicular lines have opposite and reciprocal slopes, we know we can simply flip the fraction and make it a negative to get the new slope of 3/4. Since B is the only option with that slope, we know it to be the correct answer.

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Expect a ton of trigonometry questions from this account. i know absolutely nothing

Paha777 [63]

Answer:

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Step-by-step explanation:

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2 years ago

What’s the soccer field answer to 120 yards long and 369 so do we subtract or multiply or divide

balandron [24]

Answer:

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3 years ago

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

---------------------------------------------------

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

---------------------------------------------------

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

8 0

3 years ago

Solve for x squared equals seventeen please <br>x^2 = 17

Lena [83]

Answer: 4.123 (rounded) or sqrt(17)

Step-by-step explanation: Take the square root of 17.

7 0

3 years ago

How many parallel sides can a triangl have

katrin2010 [14]

Answer:

None.

Step-by-step explanation:

Hi!!

A triangle has <u>NO</u> parrallel lines. It cannot. Think of a triangle. IF it had a parralell line(s), it would not have 3 sides.

Otherwise, it wouldn't be a polygon which is part of the criteria for a triangle.

Hope this helps!

:D

5 0

3 years ago

Read 2 more answers