URGENT HELP PLEASE!

Can someone please help me I’m struggling

blagie [28]

Answer:

point form is (2,-1), equation form is x= 2, y=-1

Step-by-step explanation:

3 0

3 years ago

Write an equation that could be used to solve for X. A Сал 2xº tº D B

m_a_m_a [10]

Answer: Hmmmm ok look i thinks its a carrot

Step-by-step explanation:

i found a x in my carrot

5 0

3 years ago

Please evaluate: (−2)^3 · (−2)^4 show your work

sammy [17]

Answer:

-128

Step-by-step explanation:

$(-2)^3 \cdot (-2)^4=$

$(-2\cdot -2\cdot -2) \cdot (-2\cdot -2\cdot -2\cdot -2)=$

$-8\cdot 16=$

$-128$

Hope this helps!

3 0

3 years ago

Read 2 more answers

The ubiquitous 12oz aluminum cans used to distribute drinks in this country have a diameter of approximately 2.75 inches and a h

gayaneshka [121]

Answer:

3.5%

Step-by-step explanation:

The volume of a cylinder = $\pi r^2h$

r = radius of cylinder,

h = height of cylinder

For the non-optimal can,

r = 2.75/2 = 1.375

h = 5.0

$V = \pi(1.375^2)\times 5.0 = 9.453125\pi$

For the optimal can,

d/h = 1,

d = h

2r = h

r = h/2

$V = \pi\left(\dfrac{h}{2}\right)^2\times h = \pi\left(\dfrac{h^3}{4}\right)$

They have the same volume.

$\pi\dfrac{h^3}{4} = 9.453125\pi$

$h^3 = 37.8125$

$h=3.36$ (This is the height of the optimal can)

$r = \dfrac{3.36}{2} = 1.68$ (This is the radius of the optimal can)

The area of a cylinder is

$A = 2\pi r(r+h)$

For the non-optimal can,

$A = 2\pi\times\dfrac{2.75}{2}\left(\dfrac{2.75}{2}+5.0\right) = 17.53125\pi$

For the optimal can,

$A = 2\pi\times1.68\left(1.68+3.36\right) = 16.9344\pi$

Amount of aluminum saved, as a percentage of the amount used to make the optimal cans = $\dfrac{17.53125\pi - 16.9344\pi}{16.9344\pi}\times 100\% = 3.5\%$

5 0

3 years ago

Help asap! Thanks in advance! There's two questions btw:)

lutik1710 [3]

Answer:

[1].

Option A and D are correct.

[2].

Option A is correct

Step-by-step explanation:

[1].

Quadratic function states that it is an equation of second degree i.,e it contains at least one term that is squared.

The standard form of the quadratic equation is; $ax^2+bx+c = 0$

A.

$y(y+4)-y = 6$

Using distributive property: $a\cdot (b+c) = a\cdot b + a\cdot c$

$y^2+4y-y=6$

Combine like terms;

$y^2+3y = 6$

or

$y^2+3y -6=0$ which represents a quadratic equation.

B.

$3a-7 = 2(7a-3)$

$3a-7 = 14a-6$

or

$11a+1 = 0$ which is not a quadratic equation.

C.

(3x+2)+(6x-1) = 0

Combine like terms;

9x +1 = 0 which is not a quadratic equation.

D.

4b(b) = 0

$4b^2 = 0$ which represents the quadratic equation.

[2].

Given the parent function: $y=x^2$

The reflection rule over x axis is given by;

$(x, y) \rightarrow (x, -y)$

then

the function become: $y = -x^2$

Vertical shift:

If c is a positive real number, the graph y=f(x)+c is the graph of y =f(x) shifted upward c units.

If c is a positive real number, the graph y=f(x)-c is the graph of y =f(x) shifted downward c units.

then;

The graph $y=-x^2-3$ is the graph of $y=-x^2$ shifted 3 units down.

Therefore, the translation of the graph of $y=x^2$ to obtain $y=-x^2-3$ is, reflect over the x-axis and shift down 3

6 0

3 years ago