If 1/2 gallon of paints cover 1/4 of a wall,then how much paint is needed to cover the entire wall?

Using the remainder theorem which quotient has a remainder of 20?

stealth61 [152]

Answer:

D

Step-by-step explanation:

When a polynomial f(x) is divided by (x - a) then remainder is f(a)

A

divided by (x + 4 ), then a = - 4

$(-4)^{4}$ - 8(- 4)² + 16 = 256 - 128 + 16 = 144 ≠ 20

B

divided by (x - 2), then a = 2

3(2)³ + 7(2)² + 5(2) + 2

= 3(8) + 7(4) + 10 + 2

= 24 + 28 + 12 = = 64 ≠ 20

C

divided by (x + 5) , then a = - 5

(- 5)³ + 5(- 5)² - 4(- 5) + 6

= - 125 + 125 + 20 + 6 = 26 ≠ 20

D

divided by (x - 2), then a = 2

3 $(2)^{4}$ - 5(2)³ + 5(2) + 2

= 3(16) - 5(8) + 10 + 2

= 48 - 40 + 12 = 20 ← Remainder of 20

3 0

2 years ago

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

c. The rocket reaches its maximum height when its vertical velocity (the $z$ -component) is 0, at which point we have

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

7 0

2 years ago

Add: (3x2 - 5x + 6) + (9 - 8x - 4x2) A) 12x2 - 13x + 2 B) 7x2 + 3x - 3 C) 7x2 - 13x + 15 D) -x2 - 13x + 15

kati45 [8]

Answer:

D

Step-by-step explanation:

So we have the expression:

$(3x^2-5x+6)+(9-8x-4x^2)$

Combine like terms:

$=(3x^2-4x^2)+(-5x-8x)+(6+9)$

Add or subtract:

$=(-1x^2)+(-13x)+(15)$

Simplify:

$=-x^2-13x+15$

Our answer is D

4 0

2 years ago

A soup can in the shape of a right circular cylinder is to be made from two materials. The material for the side of the can cost

Advocard [28]

Answer:

Radius = 1.12 inches and Height = 4.06 inches

Step-by-step explanation:

A soup can is in the shape of a right circular cylinder.

Let the radius of the can is 'r' and height of the can is 'h'.

It has been given that the can is made up of two materials.

Material used for side of the can costs $0.015 and material used for the lids costs $0.027.

Surface area of the can is represented by

S = 2πr² + 2πrh ( surface area of the lids + surface are of the curved surface)

Now the function that represents the cost to construct the can will be

C = 2πr²(0.027) + 2πrh(0.015)

C = 0.054πr² + 0.03πrh ---------(1)

Volume of the can = Volume of a cylinder = πr²h

16 = πr²h

$h=\frac{16}{\pi r^{2}}$ -------(2)

Now we place the value of h in the equation (1) from equation (2)

$C=0.054\pi r^{2}+0.03\pi r(\frac{16}{\pi r^{2}})$

$C=0.054\pi r^{2}+0.03(\frac{16}{r})$

$C=0.054\pi r^{2}+(\frac{0.48}{r})$

Now we will take the derivative of the cost C with respect to r to get the value of r to get the value to construct the can.

$C'=0.108\pi r-(\frac{0.48}{r^{2} })$

Now for C' = 0

$0.108\pi r-(\frac{0.48}{r^{2} })=0$

$0.108\pi r=(\frac{0.48}{r^{2} })$

$r^{3}=\frac{0.48}{0.108\pi }$

r³ = 1.415

r = 1.12 inch

and h = $\frac{16}{\pi (1.12)^{2}}$

h = 4.06 inches

Let's check the whether the cost is minimum or maximum.

We take the second derivative of the function.

$C"=0.108+\frac{0.48}{r^{3}}$ which is positive which represents that for r = 1.12 inch cost to construct the can will be minimum.

Therefore, to minimize the cost of the can dimensions of the can should be

Radius = 1.12 inches and Height = 4.06 inches

5 0

3 years ago

Find the distance between the points (8,1) and (2,10).<br> Round decimals to the nearest tenth.

miss Akunina [59]

The nearest distance between the points (8,1) and (2,10) is 11 units, which is mentioned in and also calculated by using formula.

Step-by-step explanation:

The given is,

Two points are (8,1) (2,10)

Step: 1

By graphical method, refer the attachment,

First point (8,1)

The values of x=8 and y=1 noted in the graph

Second point (2,10)

The values of x=2 and y=10 noted in the graph

Now join the two points (8,1) and (2,10)

Measure the distance between two points, the distance is 11 units. Due to some error the value becomes 10.98 or 11.1 we need to convert the answer into nearest whole number.

( OR )

Step:1

By formula method,

Distance = $\sqrt{(x_{2} - x_{1} )^{2} +(y_{2} - y_{1} )^{2} }$ .....................(1)