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RideAnS [48]
3 years ago
6

Find an equation of a line through (-2, 4) and perpendicular to -x+8y=16

Mathematics
2 answers:
erica [24]3 years ago
6 0

Answer: y=-8x-12

Step-by-step explanation:

The equation of the line has the form:

y=mx+b

Where m is the slope and b is the y-intercept.

Let's find the slope of the given line. Solve for y:

-x+8y=16\\8y=x+16\\y=(1/8)x+2

Then m=1/8

By definition the slopes of perpendicular lines are negative reciprocals, therefore the slope of the other line is:

m=-8

Substitute the slope and the point given into the equation of the line to calculate b:

4=(-8)(-2)+b\\4=16+b\\b=-12

Then the equation is:

y=-8x-12

Lena [83]3 years ago
3 0

Answer:

y  =  -8x-12

Step-by-step explanation:

We have given an equation of line and a point.

-x+8y=16  and (x,y) =  (-2, 4)

We have to find the equation of line that is perpendicular -x+8y=16.

y  = mx+c is equation of line where m is slope and c is y-intercept.

8y  =  x+16

y  = 1/8x+2

Hence, slope of perpendicular line is -8.

y  =  -8x+c is equation of perpendicular line to -x+8y=16.

putting the value of point in above equation , we have

4 = -8(-2)+c

4  =  16+c

c  =  4-16

c  =  -12

hence, y  = -8x-12  is equation of line passes through (-2, 4) and perpendicular to -x+8y=16.

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4 0
2 years ago
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
What is the product of (3 the square root of 8)(4 the square root of 3)? Simplify your answer
Vsevolod [243]

3\sqrt{8}\cdot4\sqrt{3} \\12\sqrt{24} \\12\cdot2\sqrt{6} \\\boxed{24\sqrt{6}}

8 0
3 years ago
What is the value of this expression when c=-4 and d = 10?<br> 1/4(c3+d2)
Mumz [18]
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

What is the value of this expression when c = -4 and d = 10 ? \frac{1}{4} (c³ + d²)

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>

<h3>Given:-</h3>

\frac{1}{4} (c³ + d²) where c = -4 and d = 10

<h3>To Find:-</h3>

The value of the expression \frac{1}{4} (c³ + d²)

<h2>Solution:-</h2>

\frac{1}{4} (c³ + d²) [Given expression]

Now, putting the value of c = -4 and d = 10 , we get,

\frac{1}{4} { (-4)³ + (10)² }

\frac{1}{4} ( -64 + 100 )

\frac{1}{4} ( 36 )

\frac{1}{4} × 36

\bf = \: 9 \: ( Answer )

3 0
3 years ago
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