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3 years ago

Which function has the largest slope?

Mathematics

1 answer:

Black_prince [1.1K]3 years ago

3 0

The answer is A. You need an explanation?

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Help me please on these questions

Fittoniya [83]

The answer to number 23 is D, 49.

8 0

3 years ago

Evaluate the expression when x=-6. x² + 5x-5

Rus_ich [418]

Answer:

Step-by-step explanation:

x² + 5x-5

Let x = -6

(-6)^2 + 5(-6) -5

36 -30 -5

6-5

6 0

3 years ago

Write an equation of the line that passes through the given points. (-1,5) and (2, - 7)

Vanyuwa [196]

Answer:

The equation of the straight line is 4x +y = 1

Step-by-step explanation:

Step(i):-

Given points are (-1,5) and ( 2,-7)

Slope of the line

$m =\frac{y_{2}-y_{1} }{x_{2} -x_{1} } =\frac{-7-5}{2-(-1)} = \frac{-12}{3} = -4$

slope of the line m = -4

Step(ii):-

The equation of the straight line passing through the point (-1,5) and having slope 'm' = -4

$y - y_{1} = m( x-x_{1} )$

y - 5 = -4 ( x-(-1))

y -5 = -4 x -4

4 x + y -5 +4=0

4x +y -1 =0

Final answer:-

The equation of the straight line is 4x +y = 1

6 0

3 years ago

F(x)=X over x^3-2x^2+5x why will this have no zeros

Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

5 0

3 years ago

How to make 155 using 3 prime numbers(You must multiply)

Citrus2011 [14]

5 is a prime factor of 155. Dividing 155 by 5, we get 31. 1*5*31 = 155.

7 0

3 years ago