If 4x³+kx²+px +2 is divisible by x²+ α prove that kp=8.

Answer:

[tex]X=6.67ft/s[/tex]

Step-by-step explanation:

From the question we are told that:

Height of pole [tex]H_p=15[/tex]

Height  of man [tex]h_m=6ft[/tex]

Speed of Man [tex]\triangle a =4ft/s[/tex]

Distance from pole [tex]d=35ft[/tex]

Let

Distance from pole to man=a

Distance from man to shadow =b

Therefore

 [tex]\frac{a+b}{15}=\frac{b}{6}[/tex]

 [tex]6a+6b=15y[/tex]

 [tex]2a=3b[/tex]

Generally the equation for change in velocity is mathematically given by

 [tex]2(\triangle a)=3(\triangle b )[/tex]

 [tex]2*4=3(\triangle b)[/tex]

 [tex]\triangle a=\frac{8}{3}[/tex]

Since

The speed of the shadow is given as

 [tex]X=\triangle b+\triangle a[/tex]

 [tex]X=4+8/3[/tex]

 [tex]X=6.67ft/s[/tex]

Answer:

the answer of r is 8 i hope it will help

Answer:

11 hours is right answer i hope it will help you

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

Explanation:

We have an octagon because there are n = 8 sides. The diagram below shows one way to number the sides so you can count them efficiently (without missing any or double counting any).

----------------

Plug n = 8 into the formula below

S = 180(n-2)

S = 180(8-2)

S = 180(6)

S = 1080

The 8 interior angles add up to 1080 degrees.

Answer:

I think you have a type.. "the seventh number must be a 1"

there are no 1's in the original set of numbers

Step-by-step explanation:

Answer:

The expected number of winning balls that Heather draws is 0.3.

Step-by-step explanation:

The balls are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Expected value of the hypergeometric distribution:

The expected value is given by:

[tex]E(X) = \frac{nk}{N}[/tex]

Expected number of blue and green balls:

40 balls, which means that [tex]N = 40[/tex]

2 are chosen, which means that [tex]n = 2[/tex]

25 are blue, which means that [tex]k = 25[/tex]

So

[tex]E(X) = \frac{nk}{N} = \frac{25(2)}{40} = 1.25[/tex]

1.25 balls are expected to be blue and 2 - 1.25 = 0.75 green.

Of the blue balls, 12% are winning.

Of the green balls, 20% are winning.

Calculate the expected number of winning balls that Heather draws.

[tex]E_w = 1.25*0.12 + 0.75*0.2 = 0.3[/tex]

The expected number of winning balls that Heather draws is 0.3.

Answer:

[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]

Step-by-step explanation:

Given

[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]

[tex]P(1) = 2[/tex]

Required

The solution

We have:

[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]

[tex]\frac{dP}{dt} = (Pt)^\frac{1}{2}[/tex]

Split

[tex]\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}[/tex]

Divide both sides by [tex]P^\frac{1}{2}[/tex]

[tex]\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}[/tex]

Multiply both sides by dt

[tex]\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt[/tex]

Integrate

[tex]\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

Rewrite as:

[tex]\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt[/tex]

Integrate the left hand side

[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt[/tex]

[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

[tex]2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

Integrate the right hand side

[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex] ---- (1)

To solve for c, we first make c the subject

[tex]c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}[/tex]

[tex]P(1) = 2[/tex] means

[tex]t = 1; P =2[/tex]

So:

[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}[/tex]

[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1[/tex]

[tex]c = 2\sqrt 2 - \frac{2}{3}[/tex]

So, we have:

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}[/tex]

Divide through by 2

[tex]P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}[/tex]

Square both sides

[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]

Answer:72 [tex]cm^{2}[/tex]

Solution 1:

Step 1: Find EF use Pythagorean theorem

[tex]EF^{2} = EB^{2} + BF^{2}[/tex]

[tex]EF^{2} = 6^{2} + 6^{2}[/tex]

EF = [tex]\sqrt{6^{2} + 6^{2} }[/tex] = 6[tex]\sqrt{2}[/tex] cm

Step 2: The area of EFGH = [tex]EF^{2}[/tex]= [tex](6\sqrt{2} )^{2}[/tex] = 72

Solution 2: See that the area of EFGH is equal [tex]\frac{1}{2}[/tex] the area of ABCD

The area of ABCD = 12x12 = 144

Thus, the area of EFGH = 144: 2 = 72:)

Have a nice day!

Answer:

A = 16 ft^2

P = 20 ft

Step-by-step explanation:

P = perimeter

A = area

STEP 1: divide the shape into rectangles

Rectangle 1: 2ft*3ft

Rectangle 2: 2ft*5ft

STEP 2: Find the area of each rectangle

Equation for area of a rectangle = bh

Rectangle 1: b = 2, h = 3

Rectangle 2: b = 2, h = 5

(2 * 3) + (2 * 5)

6 + 10

16 ft^2

Now, we have to find the perimeter

STEP 1:  Find the unknown side lengths.

To find the lengths of the sides not labeled, you have to use the lengths of the sides we already know.

The length of one parallel side is 5, and the length of another parallel side is 2. The length of the unknown side starts at the same place as the top of the side length that is 5, and ends at the top of the side length that is 2. This means that we have to subtract 2 from 5 in order to find the unknown side length.

STEP 2: Add up all the side lengths

P = 2 + 5 + 5 + 2 + 3 + 3

P = 20 ft

Don't forget to label your answers!!

I hope this made sense, it's is a little hard to explain in simple terms without being able to draw, but I hope it helped.

Answer:

7 : 8

Step-by-step explanation:

that is the procedure above

Answer:

1024

Step-by-step explanation:

Given :-

Value of -2x²y³

Answer:

254

Step-by-step explanation:

^ <- this is the square sign

-2x^y^3

x=2

y=4

put x values in to x place and y value in to y place.

-2(2)^2(4)^3

Find the squares and - it with 2

-2(4)(64)

2-256=254

:. the value of -2x^2y^3 =254

That the answer.

Hope this is what you asked.

In cylindrical coordinates, W is the set of points

W = {(r, θ, z) : 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π and -2 ≤ z ≤ 5}

(A) Then the integral of f(x, y, z) over W is

[tex]\displaystyle\iiint_W(x^2+y^2+z^2)\,\mathrm dV = \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]

(B)

[tex]\displaystyle \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta = 2\pi \int_0^2\int_{-2}^5(r^3+rz^2)\,\mathrm dz\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(zr^3+\frac13rz^3\right)\bigg|_{z=-2}^{z=5}\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(\frac{133}3r+7r^3\right)\,\mathrm dr \\\\\\= 2\pi \left(\frac{133}6r^2+\frac74r^4\right)\bigg|_{r=0}^{r=2} \\\\\\= 2\pi \left(\frac{110}3\right) = \boxed{\frac{220\pi}3}[/tex]

Answer:

AB = 5.6 cm

Step-by-step explanation:

Reference angle (θ) = 62°

Hypotenuse = 12 cm

Adjacent = AB

Apply the trigonometric ratio formula, CAH, which is:

Cos θ = Adj/Hyp

Plug in the values

Cos 62° = AB/12

12*Cos 62° = AB

5.63365876 = AB

AB = 5.6 cm (1 decimal place)

Answer:

[tex]-5 \to -24[/tex]

[tex]-1 \to -4[/tex]

[tex]2 \to 11[/tex]

[tex]3 \to 16[/tex]

[tex]4 \to 21[/tex]

Step-by-step explanation:

Given

[tex]f(x) = 5x + 1[/tex]

Required

Complete the table (see attachment)

When x = -5

[tex]f(-5) = 5 * -5 + 1 = -24[/tex]

When x = -1

[tex]f(-1) = 5 * -1 + 1 = -4[/tex]

When x = 2

[tex]f(2) = 5 * 2 + 1 = 11[/tex]

When x = 3

[tex]f(3) = 5 * 3 + 1 = 16[/tex]

When x = 4

[tex]f(4) = 5 * 4 + 1 = 21[/tex]

So, the table is:

[tex]-5 \to -24[/tex]

[tex]-1 \to -4[/tex]

[tex]2 \to 11[/tex]

[tex]3 \to 16[/tex]

[tex]4 \to 21[/tex]

Answer:

[tex]\bar x = 107.11[/tex]

[tex]\sigma_x = 31.07[/tex]

Step-by-step explanation:

See comment for complete question

Given

[tex]x: 97\ 178\ 129\ 90\ 75\ 94\ 116\ 100\ 85[/tex]

Solving (a): The sample mean

This is calculated using:

[tex]\bar x = \frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x = \frac{97+ 178+ 129+ 90+ 75+ 94+ 116+ 100+ 85}{9}[/tex]

[tex]\bar x = \frac{964}{9}[/tex]

[tex]\bar x = 107.11[/tex]

Solving (b): The sample standard deviation

This is calculated as:

[tex]\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]

So, we have:

[tex]\sigma_x = \sqrt{\frac{(97 - 107.11)^2 +.............+ (85- 107.11)^2 }{9-1}}[/tex]

[tex]\sigma_x = \sqrt{\frac{7720.8889}{8}}[/tex]

[tex]\sigma_x = \sqrt{965.1111125}[/tex]

[tex]\sigma_x = 31.07[/tex]

Answer:

-  [tex]\frac{4}{\sqrt{29} }[/tex]

Step-by-step explanation:

The equations for the 1st object :

x₁ = 10 cos(2t),  and  y₁ = 6 sin(2t)

2nd object :

x₂ = 4 cos(t), y₂ = 4 sin(t)

Determine rate at which distance between objects will continue to change

solution Attached below

Distance( D )  = [tex]\sqrt{(10cos2(t) - 4cos(t))^2 + (6sin2(t) -4sin(t))^2}[/tex]

hence; dD/dt = - [tex]\frac{4}{\sqrt{29} }[/tex]

Answer:

{-6,0,2,4}

Step-by-step explanation:

9514 1404 393

Answer:

  y -9

Step-by-step explanation:

From 4 years ago until 5 years from now, John will age 9 years. That is, his age 4 years ago is 9 years less than it will be in 5 years.

John's age 4 years ago is y-9 years.

Answer:

-1.28 AND 2.61

Step-by-step explanation:

[tex]10= -4x+3x^2\\ 3x^2 -4x - 10 = 0\\\\[/tex]

use quadratic formula

x =  [tex]\frac{-b+\sqrt{b^{2} -4ac} }{2a}[/tex]   x =  [tex]\frac{-b-\sqrt{b^{2} -4ac} }{2a}[/tex]

Solution/X-Intercepts: -1.28 AND 2.61    

Answer:

in a square all sides are equal so x has to equal

128

Hope This Helps!!!

im struggling with the same one

Answer:

Step 1

Because the number in front of the bracket is 1 and it is also affected by the negative sign(-),5 is supposed to be negative not positive because (negative by positive is negative)

And since the first step has an error in it,the remaining steps would also be wrong.

Answer:

hey hi mate

hope you like it

plz mark it as brainliest

Answer:

10

Step-by-step explanation:

Sorry. I needed to answer this question to get access.

[tex]\frac{13}{28}[/tex]

Given:

Blue marbles: 3

Reb marbles: 5

Total marbles: 8

Two marbles are selected at random, one after the other with replacement.

Getting the same colour of marbles from the selection means the two marbles are both red or both blue.

(a) Probability of getting 2 marbles being red in colour

i. Probability of picking a red at the first selection:

Number of red marbles ÷ Total number of marbles

=> 5 ÷ 8 = [tex]\frac{5}{8}[/tex]

ii. Probability of picking a red at the second selection:

Number of remaining red marbles ÷ Total number of remaining marbles

Since after the first pick, the marble is not replaced, the remaining red marbles is 4 while the total number of remaining marbles is 7

=> 4 ÷ 7 = [tex]\frac{4}{7}[/tex]

iii. The probability of getting both marbles being red is the product of i and ii above. i.e

[tex]\frac{5}{8}[/tex] x [tex]\frac{4}{7}[/tex] = [tex]\frac{5}{14}[/tex]

(b) Probability of getting 2 marbles being blue in colour

i. Probability of picking a blue at the first selection:

Number of blue marbles ÷ Total number of marbles

=> 3 ÷ 8 = [tex]\frac{3}{8}[/tex]

ii. Probability of picking a blue at the second selection:

Number of remaining blue marbles ÷ Total number of remaining marbles

Since after the first pick, the marble is not replaced, the remaining blue marbles is 2 while the total number of remaining marbles is 7

=> 2 ÷ 7 = [tex]\frac{2}{7}[/tex]

iii. The probability of getting both marbles being blue is the product of i and ii above. i.e

[tex]\frac{3}{8}[/tex] x [tex]\frac{2}{7}[/tex] = [tex]\frac{3}{28}[/tex]

(c) Probability of getting 2 marbles of the same colour.

The probability of getting 2 marbles of same colour is the sum of the probability of getting both marbles of red colour and the probability of getting both marbles as blue colour. i.e The sum of a(iii) and b(iii)

[tex]\frac{5}{14}[/tex] + [tex]\frac{3}{28}[/tex] = [tex]\frac{13}{28}[/tex]

The probability of getting 2 of the same colour is [tex]\frac{13}{28}[/tex]

Answer:

Step-by-step explanation:

19.

The numbers are m and n

Sum of m and n = m + n

Sum is multiplied by 91 = 91 x ( m + n )

20.

Let the number be = m

Six subtracted from the number = m - 6

41 times the difference = 41 x ( m - 6)

21.

Let the number be = r

Difference between 83 and 10 = 83 - 10 = 73

[tex]The \ number\ divided \ by\ the \ difference \ = \frac{r}{73}[/tex]

22.

Total of p and 23 = p + 12

[tex]Total \ divided \ by \ 18 = \frac{p + 12 }{18}[/tex]

23.

The product of c and 3  = 3c

Sum of 9 and 12 = 21

Product is more than Sum = 3c + 21

24.

Sum of y  and 10 = y + 10

Number x decreased by 5 = x - 5

[tex]Sum \ divided \ by \ difference = \frac{ y + 10 }{x - 5}[/tex]

Answer:

annualy=$3689.62

semiannually=$3695.27

monthly=$3700.06

weekly=$3700.81

daily=$3701.00

Continuously=$3701.03

Step-by-step explanation:

Given:

P=3000

r=3%

t=7 years

Formula used:

Where,

A represents Accumulated amount

P represents (or) invested amount

r represents interest rate

t represents time in years

n represents accumulated or compounded number of times per year

Solution:

(i)annually

n=1 time per year

[tex]A=3000[1+\frac{0.03}{1} ]^1^(^7^)\\ =3000(1.03)^7\\ =3689.621596\\[/tex]

On approximating the values,

A=$3689.62

(ii)semiannually

n=2 times per year

[tex]A=3000[1+\frac{0.03}{2}^{2(4)} ]\\ =3000[1+0.815]^14\\ =3695.267192[/tex]

On approximating the values,

A=$3695.27

(iii)monthly

n=12 times per year

[tex]A=3000[1+\frac{0.03}{12}^{12(7)} \\ =3000[1+0.0025]^84\\ =3700.0644[/tex]

On approximating,

A=$3700.06

(iv) weekly

n=52 times per year

[tex]A=3000[1+\frac{0.03}{52}]^3^6 \\ =3000(1.23360336)\\ =3700.81003[/tex]

On approximating,

A=$3700.81

(v) daily

n=365 time per year

[tex]A=3000[1+\frac{0.03}{365}]^{365(7)} \\ =3000[1.000082192]^{2555}\\ =3701.002234[/tex]

On approximating the values,

A=$3701.00

(vi) Continuously

[tex]A=Pe^r^t\\ =3000e^{\frac{0.03}{1}(7) }\\ =3000e^{0.21} \\ =3000(1.23367806)\\ =3701.03418\\[/tex]

On approximating the value,

A=$3701.03

The work is simply the dot product of the force and displacement (which I assume are given in Newtons and meters, respectively):

W = F • d

W = (5i + 3j - 2k) N • ((4i - j + 3k) m - (j + k) m)

W = (5i + 3j - 2k) • (4i - 2j + 2k) Nm

W = (20 - 6 - 4) Nm

W = 10 J