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yanalaym [24]
3 years ago
6

(1,5) with a slope of -2

Mathematics
2 answers:
galina1969 [7]3 years ago
5 0

answer:

Linear equation :

y= -2x +7

Step-by-step explanation:

laiz [17]3 years ago
3 0

Answer:

y = -2x + 7 is the equation if that's what you are asking for.

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Aliun [14]
N=199 because in order to get a a smaller negative you would have to add a positive into a negative
6 0
3 years ago
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Simplify x + 4 + 3x
Ostrovityanka [42]
X+4+3x =
4x+4 (you sum the like terms )
3 0
3 years ago
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The equation of the line CD is (y-3)=-2(x-4). what is the scope of the line perpendicular to line CD
Delicious77 [7]

Answer:

y=-1/2x-5

Step-by-step explanation:

First we need to put (y-3)=2(x-4) in slope intercept form.

we need to do order of operations....

so we need to first do the parenthesis

x*2= 2x

2*-4= -8

y-3=2x-8

add 3 on both sides to isolate y

y= 2x-5

so now we know the y=mx+b form

this is one line...

now we need to find out what line is perpendicular.

Perpendicular lines have opposite reciprocals slopes.

y=-1/2x-5

8 0
3 years ago
Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify
irina [24]

Answer:

x-coordinates of relative extrema = \frac{-1}{2}

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}

Differentiate with respect to x

f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}

f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}

Differentiate f'(x) with respect to x

f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1

At x = \frac{-1}{2},

f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0

We know that if f''(a)>0 then x = a is a point of minima.

So, x=\frac{-1}{2} is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0
3 years ago
-2m+8+m+1=0 <br>What does m equal?
Sergio [31]

Answer:

m = 9

Step-by-step explanation:

-2m + 8 + m + 1 = 0

-m + 9 = 0

9 = m

3 0
3 years ago
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