All categories

3 years ago

Identify the equation in slope-intercept form for the line with the given slope that contains the given point. Slope =2; (1,4)

1 answer:

3 0

Slope-intercept form: y=mx+b, with m being the slope and b being the y-intercept. First, plug in the given slope into the slope form:

y=2x+b

Now, given the coordinates (1,4), you can plug the coordinates into the equation for x and y respectively, to solve for b (the y-intercept).

4=2(1)+b

4=2+b

Subtract 2 from both sides.

2=b

The y-intercept is 2.

Now, plug that into the equation:

y=2x+2

The answer is y=2x+2.

I hope this helps :)

You might be interested in

Justin made $204 for 12 hours of work. At the same rate, how much would he make for 18 hours of work?

lianna [129]

$306. He makes $17 an hour

4 0

3 years ago

Make groups of ten. Write the tens and ones. 1) 90 Ones

Alisiya [41]

90 ones would make 9 tens. It makes 9 tens because theres a 0 after 90. The 0 represents another place value. You can write this as 9 groups of ten

7 0

3 years ago

#18 how would u not multiplying to find which is grater

Snowcat [4.5K]

In the first product the numerator of the fraction is greater than the denominator, therefore the fraction is greater than 1. On the other hand in the second product the numerator of the fraction is smaller than the denominator, therefore the fraction is smaller than one. That means the first product is greater than the second one.

3 0

3 years ago

Read 2 more answers

Write a polynomial equation with integer coefficients that has the given roots x=1, x=-4

Gekata [30.6K]

(x -1)(x +4) = 0

x² +3x -4 = 0 . . . . . . probably the form you're looking for

7 0

3 years ago

Read 2 more answers

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$

This is what the Bilateral Theorem says:

$\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L$

4 0

4 years ago