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3 years ago

Can someone help me with this problem? Please.

Mathematics

1 answer:

Mamont248 [21]3 years ago

4 0

Answer:

(7x-1)-3(7x-1)

Step-by-step explanation:

First write out the problem

7x^2-22x+3

Next, Factor the expressions

7x^2 - x - 21x + 3

Now factor the whole thing

x(7x-1)-3(7x-1)

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Angle AOC has what measurement according to the protractor? A. 150° B. 50° C. 30° D. 130°

maxonik [38]

Answer:

∠AOC = 50°

B is correct.

Step-by-step explanation:

In the given figure of protractor measure the ∠AOC

Please find the attachment of protractor.

OA is horizontal line which is base of protractor.

In angle AOC, two legs OA and OC

OA is base of protractor. OC another leg.

Now we see position of leg OC of angle AOC.

∠AOC is acute angle. So, see the number bottom of protractor.

Hence, the measure of ∠AOC is 50°

7 0

3 years ago

Read 2 more answers

Find the missing angle.

Vika [28.1K]

Answer:

X = 45

Step-by-step explanation:

Hope it will be helpful for you.

8 0

2 years ago

In Exercise,find the horizontal asymptote of the graph of the function. f(x) = 8x^3+2/2x^3+x

KIM [24]

Answer:

Horizontal asymptote of the graph of the function f(x) = (8x^3+2)/(2x^3+x) is at y=4

Step-by-step explanation:

I attached the graph of the function.

Graphically, it can be seen that the horizontal asymptote of the graph of the function is at y=4. There is also a vertical asymptote at x=0

When denominator's degree (3) is the same as the nominator's degree (3) then the horizontal asymptote is at (numerator's leading coefficient (8) divided by denominator's lading coefficient (2)) $y=\frac{8}{2}=4$

6 0

3 years ago

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

Can someone help me solve this

Scrat [10]

Answer:

your photo dose not work

Step-by-step explanation:

but you would half to email it to me then i can help you if you want to do that

8 0

3 years ago