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2 years ago

Ind the slope m of the tangent to the curve y = 9 + 4x2 − 2x3 at the point where x =

1 answer:

3 0

Find the derivative:-
y' = 8x - 6x^2
This is the slope in terms of x.
When x = a , the slope is
8a - 6a^2

(b) equation of tangent line at x1,y1
is y- y1 = (8x - 6x^2)( x - x1) , so at (1,11) it is
y - 11 = (8(1) - 6((1)^2) ) (x - 1)
y = 2(x - 1) + 11
y = 2x + 9 (answer)

At (2,9)
y - 9 = (16-24)(x - 2)
y - 9 = -8x + 16
y = -8x + 25 answer

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How much more expensive is it, per pound, to buy ground beef at Store B than at Store A?

Ivahew [28]

Answer:

2.42

Step-by-step explanation:

First you need to use two points from store b to find the slope (y1-y2)/(x1-x2). I chose the first two points. (15.54-25.9)/(1.5-2.5)= 10.36. After you take another point from store b to plug into the equation y1-y2=m (x1-x2). M is the slope we just found and I used the first point.

Y1-15.54=10.36 (x1-1.5) distribute the 10.36 to the parentheses.

Y1-15.54=10.36x -15.54 get y1 by itself

Y=10.36x so store b is 10.36 a pound and store a is 7.94 a pound. 10.36-7.94= 2.42

7 0

2 years ago

(06.04 MC)

Andru [333]

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

◉ $\large\bm{ -4}$

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

Before performing any calculation it's good to recall a few properties of integrals:

$\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}$

$\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }$

So we apply the first property in the first expression given by the question:

$\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}$

And we solve the second integral:

$\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }$

Then we take the last equation and we subtract 10 from both sides:

$\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}$

And we divide both sides by 2:

$\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}$

Then we apply the second property to this integral:

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}$

Then we use the other equality in the question and we get:

$\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}$

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}$

We substract 8 from both sides:

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}$

• $\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}$

7 0

1 year ago

Question 1<br> Let f(x) = 2x^2 - 3x - 4 and g(x) = x + 5. Find (fg)(x).

Scilla [17]

The function (fg)(x) is a composite function

The value of the function (fg)(x) is 2x^3 + 7x^2 - 19x - 20

<h3>How to determine the function (fg)(x)?</h3>

The functions are given as:

f(x) = 2x^2 - 3x - 4 and g(x) = x + 5.

To calculate (fg)(x), we make use of

(fg)(x) = f(x) * g(x)

So, we have:

(fg)(x) = (2x^2 - 3x - 4) * (x + 5)

Expand

(fg)(x) = 2x^3 - 3x^2 - 4x + 10x^2 - 15x - 20

Collect like terms

(fg)(x) = 2x^3 - 3x^2 + 10x^2 - 4x - 15x - 20

Evaluate

(fg)(x) = 2x^3 + 7x^2 - 19x - 20

Hence, the function (fg)(x) is 2x^3 + 7x^2 - 19x - 20

Read more about composite function at:

brainly.com/question/10687170

7 0

2 years ago

The length of the edge of a cube is ³√7 cm. Find the volume of a cube

slava [35]

Answer:

7 $cm^{3}$

Step-by-step explanation:

volume is length X width X height

since a cube has the same measurement for length, width, and height, the volume is found by multiplying $\sqrt[3]{7}$ X $\sqrt[3]{7}$ X $\sqrt[3]{7}$

the answer is 7

8 0

3 years ago

at McDonalds a cheese burger, c, has 200 fewer calories that a large fry, f. two cheeseburgers and a large fry have 1100 calorie

lora16 [44]

Answer:

700 more caleries

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers