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2 years ago

The graph of the function, f(x) = - x2 - 3x + 1, opens and has a, value.

Mathematics

2 answers:

faust18 [17]2 years ago

8 0

Answer:

Opens downwards and has a maximum value

Step-by-step explanation:

f(x) = -x² - 3x + 1

a is the coefficient of x², which is -1

Since a < 0, it's a concave down graph which opens downwards. Therefore it has a maximum turning point

frosja888 [35]2 years ago

6 0

Answer:

downward, maximum value

Step-by-step explanation:

f(x) = - x^2 - 3x + 1

The negative in front of the x^2 means the graph opens downwards

The y intercept is 1

Because it opens downward it has a maximum value

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The sum of Eric’s and Bob’s weights is 9 times the difference of their weights. The positive difference of their weights is also

Pie

Answer:

150

Step-by-step explanation:

So I will shorten Eric, and Bob's names as E, and B.

So the equations are this

B-E = B+E - 240, E as Eric, and B as Bob, and this E+B=9(E-B)

Move the variables and you get

-2E =- 240

Just divide them both by -2

and E=120

So when we know the value of E we can just plug it into the 1st question

120+B=9B+1080

Moves the Variables and numbers to the other side

8B=1200

1200/8= 150

B=150

There is Bob's weight

6 0

1 year ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

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

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

1) A medium cookies and creme milkshake at The

Leya [2.2K]

Answer:

You will pay $4.31

Step-by-step explanation:

I know this because a fourth of $5.75 is 1.4375 so you would round it to $1.44 and then $5.75 - $1.44 is $4.31.

So you will pay $4.31 for one milkshake.

5 0

2 years ago

If X²⁰¹³ + 1/X²⁰¹³ = 2, then find the value of X²⁰²² + 1/X²⁰²² = ?

enyata [817]

Step-by-step explanation:

$\bf➤ \underline{Given-} \\$

$\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\$

$\bf➤ \underline{To\: find-} \\$

$\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\$

$\bf ➤\underline{Solution-} \\$

Let us assume that:

$\rm: \longmapsto u = {x}^{2013}$

Therefore, the equation becomes:

$\rm: \longmapsto u + \dfrac{1}{u} = 2$

$\rm: \longmapsto \dfrac{ {u}^{2} + 1}{u} = 2$

$\rm: \longmapsto{u}^{2} + 1 = 2u$

$\rm: \longmapsto{u}^{2} - 2u + 1 =0$

$\rm: \longmapsto {(u - 1)}^{2} =0$

$\rm: \longmapsto u = 1$

Now substitute the value of u. We get:

$\rm: \longmapsto {x}^{2013} = 1$

$\rm: \longmapsto x = 1$

Therefore:

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 1 + 1$

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 2$

★ Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3ab(a - b) - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x - b) = x² + (a - b)x - ab

(x - a)(x + b) = x² - (a - b)x - ab

(x - a)(x - b) = x² - (a + b)x + ab

6 0

2 years ago

the length of a rectangle is 2 cm less than five times the width. the area of the rectangle is 16 cm^2. find the length and widt

ahrayia [7]

length = L
width = w

L = (5w-2) cm

$w(5w-2)=16 \\ 5w^2-2w-16=0 \\ D=b^2-4ac=(-2)^2-4*5*(-16)=324 \\ w_{1,2}= \frac{-bб \sqrt{D} }{2a} \\ w_1= \frac{2- \sqrt{324} }{2*5}= \frac{2-18}{10}=-1.6 \ \ \O \\ w_2= \frac{2+18}{10}=2 cm$

width = 2 cm
length = 5w-2 = 5*2-2 = 8 cm

8 0

3 years ago