All categories

3 years ago

At how many points does the graph of the function below intersect the x-axis?

Mathematics

2 answers:

Ne4ueva [31]3 years ago

5 0

Answer:

A.0

Step-by-step explanation:

y=4x²- 9x + 9

The function will intersect the x-axis when y = 0.

4x² - 9x + 9 = 0

a=4, b= -9, c=9

D=b² - 4ac = (-9)²- 4*4*9=81 - 144= - 63

D= -63, D<0, so the equation

4x² - 9x + 9 = 0 does not have real roots. That means that function y=4x²- 9x + 9 does not intersect x- axis. So, answer is A. 0.

antoniya [11.8K]3 years ago

5 0

Answer:

Step-by-step explanation:

Apex

You might be interested in

99 POINTS AND BRAINLIST!!!! BE FIRST WITH THE "RIGHT" ANSWER!!

VladimirAG [237]

Answer:

The answer should be 150 x .20 = 30m/s

5 0

3 years ago

Read 2 more answers

3⁷×3-⁴ I need help asap!

SCORPION-xisa [38]

Hope it helps

Have a good day Ahead

8 0

2 years ago

Read 2 more answers

Hello people ~

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

Using Pythagoras Theorem

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

First derivative

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

apply chain rule

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

Equate the first derivative to zero, that is V'(x) = 0

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers

Please make Y=x+12 and 4x+2y=27 into one equation.

tiny-mole [99]

So what you need to start with is taking the x from both sides as adding 12 to them. Then, you move the 27 over 4 decimals.

4 0

2 years ago

raheel travelled 3654 km by bus and train. if he went 2365 km by bus, how far did he travel by train?

gladu [14]

3654 - 2365 = 1289 km

4 0

3 years ago

Read 2 more answers