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3 years ago

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Function L is represented by the equation y 6x +4. Compare Functions B an determining which one has the greater rate of change a

nd which one has the gie y-intercept. Explain why your answers are correct. Show your work.please

2 answers:

Alenkasestr [34]3 years ago

7 0

Answer:

-1

Step-by-step explanation:

Eva8 [605]3 years ago

6 0

Answer:

Both the rate of change and the y-intercept are greater in function L

Step-by-step explanation:

To find which has the greater rate of change and the greater y-intercept, start by modeling the equation in the table. We can do that by first finding the slope using two points and the slope formula.

m(slope) = (y2 - y1)/(x2 - x1)

m = (5 - -1)/(1 - -1)

m = 6/2

m = 3

Now using that slope and a point, we can find the equation using point-slope form.

y - y1 = m(x - x1)

y - 5 = 3(x - 1)

y - 5 = 3x - 3

y = 3x + 2

Now we can compare the two equations. First, we know the rate of change is greater in equation L due to the coefficient of x being higher. We also know that the y-intercept is greater due to the constant at the end being greater.

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Pls, help with this math question

Aleks04 [339]

Step-by-step explanation:

OCB=CBO(base angle of isosceles triangle are equal)

now,

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BOC=180-62

BOC=118°

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<h2>stay safe healthy and happy...</h2>

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3 years ago

Which of the following is equivalent to 1 cup

kipiarov [429]

1 cup=16 tablespoons

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2 years ago

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Find the value of f(-9)

almond37 [142]

Answer: 2

Step-by-step explanation:

When x=-9, y=2. So, f(-9)=2.

7 0

2 years ago

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Leni [432]

Answer:

a)

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

b)

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on that line.

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$ .

So we know $(x_1,y_1)=(3,-5)$ .

To find $m$ , we must calculate the derivative of $g$ at $x=3$ :

$m=g'(3)=(3)^2+7=9+7=16$ .

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$ .

I'm going to solve this for $y$ .

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$ .

Let's evaluate this approximation function for $g(2.9)$ .

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$ .

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$ .

$2x$ is negative for $x$ .

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$ .

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.

6 0

3 years ago

Giving 20 points and brainiest!

sladkih [1.3K]

Answer:

1/2,3/6,5/10,8/16

Step-by-step explanation:

they all equal 1/2

Im not sure if thats what it was asking but yea

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2 years ago

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