find the total surface area and volume, then multiply by 0.0004

The solutions to the questions are given below

a)

sample(n) word length sample mean

1                    5,4,4,2    3.75

2                    3,2,3,6     3.5

3                         5,6,3,3    4.25

b)R =0.75

c)

The table below shows sample means in the table.

sample(n) word length sample mean

1                    5,4,4,2    3.75

2                    3,2,3,6     3.5

3                         5,6,3,3    4.25

b)

Generally, the equation for is  mathematically given as

variation in the sample means

R =maximum -minimum

R=4.25-3.5

R =0.75

c)

In conclusion, In most cases, the mean of many samples will provide a more accurate estimate than the mean of a single sample.

They may have a higher level of confidence in their estimate if the range of the sample means is closer to 0 than it is now.

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Answer:

3x + 16 + 30/x - 13

Step-by-step explanation:

From the two functions function 2 has the highest maximum value.

Given two functions,one is y=-[tex]x^{2} -2x-2[/tex] and f(-2)=1,f(-1)=6,f(0)=9,f(1)=10,f(2)=9,f(3)=6.

We ae required to choose a function which is having highest maximum possible value.

Function is like a relationship between two or more variables expressed in equal to form. Each value of x of a function has some value of y.

The highest value in the second function is 10 which is at x=1. Put the value of x=1 in y=-[tex]x^{2} -2x-2[/tex].

y=-1-2-2

y=-5

So it might not be highest value of this function.So the second function has highest maximum value of 10 at x=1.

Hence the second function has highest value.

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[tex] \sf \blue{True✅}[/tex]

Yes, it's true sometimes we don't use negative value of 'x'.

So, we conclude it's True.

Answer:

a. triangle DEF's lines are all 3 times more than triangle ABC

b. triangle DEF's lines are all 3 times more than triangle ABC

c. 2.5 times 2 = 5

I hope u can elaborate a bit more :)

Answer:

-8

Step-by-step explanation:

y=(1/3)x

4 = (1/3)12

y=(1/3)(-24)

y= -8

There is only one solution for the equation 4z + 2(z -4) = 3z + 11 because the exponent for the power of z is 1.

In mathematics, an equation is a formula that expresses the equality of two expressions, by connecting them with the equals sign =.

A solution is any value of a variable that makes the specified equation true.

According to the given information:

4z + 2(z-4)= 3z+11

Solve the equation,

4z+2z-8=3z+11

6z-3z=11+8

3z =19

z=

Hence,

Number of solution that can be found for the equation 4z + 2(z-4)= 3z+11 is option(2) one

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Answer:

OA=95

Step-by-step explanation:

JL = 19Z = OA = 4Z+75

19Z = 4Z+75

19Z-4Z=75

15Z=75

Z=75÷15

Z=5

OA=4(5)+75

OA=95

btw, u stan blackpink? i cant wait for their next comeback "BORN PINK"

Answer:

one hour

Step-by-step explanation:

if they are going the same speed going differnet directions they will be 2 miles apart after one hour

Answer:

1441.503 square inches

Step-by-step explanation:

155×60cm=9300square centimeter

9300 divide 6.452 to change to square inches.

Ans=1441.503

Answer:

64.58 sq.in

Step-by-step explanation:

Answer:

[tex]\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=-4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

[tex]\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x[/tex]

Factor the denominator:

[tex]\begin{aligned}\implies x^3-2x^2+x & = x(x^2-2x+1)\\& = x(x^2-x-x+1)\\& = x(x(x-1)-1(x-1))\\ & = x((x-1)(x-1))\\& = x(x-1)^2\end{aligned}[/tex]

[tex]\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int \dfrac{x^2-4}{x(x-1)^2}\:\:\text{d}x[/tex]

Take partial fractions of the given fraction by writing out the fraction as an identity:

[tex]\begin{aligned}\dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A}{x}+\dfrac{B}{(x-1)}+\dfrac{C}{(x-1)^2}\\\\ \implies \dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A(x-1)^2}{x(x-1)^2}+\dfrac{Bx(x-1)}{x(x-1)^2}+\dfrac{Cx}{x(x-1)^2}\\\\ \implies x^2-4 & \equiv A(x-1)^2+Bx(x-1)+Cx \end{aligned}[/tex]

Calculate the values of A and C using substitution:

[tex]\textsf{when }x=0 \implies -4=A(1)+B(0)+C(0) \implies A=-4[/tex]

[tex]\textsf{when }x=1 \implies -3=A(0)+B(0)+C(1) \implies C=-3[/tex]

Therefore:

[tex]\begin{aligned}\implies x^2-4 & \equiv -4(x-1)^2+Bx(x-1)-3x\\& \equiv -4(x^2-2x+1)+B(x^2-x)-3x\\& \equiv -4x^2+8x-4+Bx^2-Bx-3x\\& \equiv (B-4)x^2+(5-B)x-4\\\end{aligned}[/tex]

Compare constants to find B:

[tex]\implies 1=B-4 \implies B=5[/tex]

Substitute the found values of A, B and C:

[tex]\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-\dfrac{3}{(x-1)^2}\:\:\text{d}x[/tex]

[tex]\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}[/tex]

[tex]\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating} $\dfrac{1}{x}$\\\\$\displaystyle \int \dfrac{1}{x}\:\text{d}x=\ln |x|+\text{C}$\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating $ax^n$}\\\\$\displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C}$\end{minipage}}[/tex]

[tex]\begin{aligned}\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x & =\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4\int \dfrac{1}{x}\:\:\text{d}x+5\int \dfrac{1}{(x-1)}\:\:\text{d}x-3 \int (x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x\end{aligned}[/tex]

Use Integration by Substitution:

[tex]\textsf{Let }u=(x-1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}[/tex]

Therefore:

[tex]\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x[/tex]

[tex]\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int u^{-2}}\:\:\text{d}u[/tex]

[tex]\implies -4 \ln |x|+5 \ln|x-1|-\dfrac{3}{-1}u^{-2+1}+\text{C}[/tex]

[tex]\implies -4 \ln |x|+5 \ln|x-1|+3u^{-1}+\text{C}[/tex]

[tex]\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{u}+\text{C}[/tex]

Substitute back in u = (x - 1):

[tex]\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}[/tex]

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[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

A linear function is an equation of the form:

[tex]\sf{y=mx+b}[/tex]

▪ [tex]\sf{m= \: slope}[/tex]

▪ [tex] \sf{b= \: y-intercept}[/tex]

First, pick any two pairs of points from the table.

[tex]\small\longrightarrow \sf{(x_1,y_1)(-1.13)}[/tex]

[tex]\small\longrightarrow \sf{(x_2,y_2)=(0,10)}[/tex]

Using these points, find the slope.

[tex]\small\longrightarrow \sf{m= \dfrac{y _{2 } - y_1}{x_2-y_1} }[/tex]

When x=0, y=10, therefore, the y-intercept, b=10.

[tex]\leadsto[/tex] Substitute m=-3 and b=10 into the slope-intercept form given above:

[tex]\small\longrightarrow \sf{m=m = \dfrac{10 - 13 }{0 -( - 1) } = - \dfrac{3}{1} = - 3}[/tex]

▪ [tex]\small\longrightarrow \sf{y= -3x+10}[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

[tex]\large\boxed{\bm{y= -3x+10}}[/tex]

The two possible positions of point P are:

(3, -5) and (0.6, 5.8)

We know that point P lies on the line:

y = 4 - 3*x

And that the distance between P and the origin is √34, then if the coordinates of point P are (x, y), we have that:

[tex]\sqrt{34} = \sqrt{x^2 + y^2}[/tex]

Now, we can replace "y" in the distance equation by the linear equation, and also remove the square roots:

[tex]34 = x^2 + (4 - 3x)^2[/tex]

Now we can solve the quadratic equation for x:

[tex]34 = x^2 + 9x^2 + 16 - 24x\\\\10x^2 - 24x - 18 = 0\\\\[/tex]

The solutions are:

[tex]x = \frac{24 \pm \sqrt{(-24)^2 - 4*10*(-18)} }{2*10} \\\\x = \frac{24 \pm36}{20}[/tex]

So the two solutions are:

x = (24 + 36)/20 = 3

x = (24 - 36)/20 = -0.6

To get the points, we need to evaluate y on these values:

y = 4 - 3*3 = -5   So we have P = (3, -5)

y = 4 - 3*(-0.6) = 2.2 So we have the point (0.6, 5.8)

There are the two possible positions of point P.

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The numeric value of the expression -a² - 2bc - |c| for a = -3, b = -5 and c = 2 is of 9.

The numeric value of an expression is found replacing each letter by it's attributed value.

In this problem, the expression is:

-a² - 2bc - |c|

The attributed values are:

a = -3, b = -5 and c = 2

Hence the numeric value will be given by:

-a² - 2bc - |c| = -(-3)² - 2(-5)(2) - |2| = -(9) + 20 - 2 = -9 + 18 = 9.

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The definition of the given piecewise function is:

A piecewise function is a function that has different definitions, depending on the input.

In this problem, for x until 2, the function is the cube function shifted down 3 units, hence the definition is:

[tex]y = x^3  - 3, x \leq 2[/tex]

For x greater than 2, the function is the square function shifted up 6 units, hence the definition is:

[tex]y = x^2 + 6, x > 2[/tex]

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[tex]denote \: by \: s \: the \: cost \: of \: a \: sweatshirt \\ denote \: by \: t \: the \: cost \: of \: a \: t \: shirt[/tex]

[tex]3t + 7s = 240.5 \\ 6t + 5s = 220[/tex]

[tex] - 6t - 14s = - 481 \\ 6t + 5s = 220[/tex]

[tex] - 9s = - 261 \\ s = \frac{ - 261}{ - 9} = 29 \: dollars[/tex]

[tex]3t + 7(29) = 240.5 \\ [/tex]

3t + 203 = 240.5

3t = 37.5

t = 12.5 dollars

Answer:

300

Step-by-step explanation:

375 divided by 5=75

Each “1”=75

4x75=300

There are 300 boys.

Hope this makes sense

Answer:

300 boys

Step-by-step explanation:

Let the number of boys = x

Girls: 375

boys : girls = 4 : 5

total in the ratio is 9

girls are 5/9 of total, and are 375

boys are 4/9 of total

5/9 x = 375

x = 375 × 9/5

x = 675

Boys are 4/9 of total.

4/9 × 675 = 300

Let [tex]X(s)[/tex] and [tex]Y(s)[/tex] denote the Laplace transforms of [tex]x(t)[/tex] and [tex]y(t)[/tex].

Taking the Laplace transform of both sides of both equations, we have

[tex]\dfrac{dx}{dt} + 3x + \dfrac{dy}{dt} = 1 \implies \left(sX(s) - x(0)\right) + 3X(s) + \left(sY(s) - y(0)\right) = \dfrac1s \\\\ \implies (s+3) X(s) + s Y(s) = \dfrac1s[/tex]

[tex]\dfrac{dx}{dt} - x + \dfrac{dy}{dt} = e^t \implies \left(sX(s) - x(0)\right) - X(s) + \left(sY(s) - y(0)\right) = \dfrac1{s-1} \\\\ \implies (s-1) X(s) + s Y(s) = \dfrac1{s-1}[/tex]

Eliminating [tex]Y(s)[/tex], we get

[tex]\left((s+3) X(s) + s Y(s)\right) - \left((s-1) X(s) + s Y(s)\right) = \dfrac1s - \dfrac1{s-1} \\\\ \implies X(s) = \dfrac14 \left(\dfrac1s - \dfrac1{s-1}\right)[/tex]

Take the inverse transform of both sides to solve for [tex]x(t)[/tex].

[tex]\boxed{x(t) = \dfrac14 (1 - e^t)}[/tex]

Solve for [tex]Y(s)[/tex].

[tex](s - 1) X(s) + s Y(s) = \dfrac1{s-1} \implies -\dfrac1{4s} + s Y(s) = \dfrac1{s-1} \\\\ \implies s Y(s) = \dfrac1{s-1} + \dfrac1{4s} \\\\ \implies Y(s) = \dfrac1{s(s-1)} + \dfrac1{4s^2} \\\\ \implies Y(s) = \dfrac1{s-1} - \dfrac1s + \dfrac1{4s^2}[/tex]

Taking the inverse transform of both sides, we get

[tex]\boxed{y(t) = e^t - 1 + \dfrac14 t}[/tex]

The scale factor is 2.875 miles per inch. Then the actual distance from the library to the park will be 2.875 miles.

Dilation is the process of increasing the size of an item without affecting its form. Depending on the scale factor, the object's size can be raised or lowered. There is no effect of dilation on the angle.

Micah drew a map of his neighborhood. The actual distance from his house to the school is 5.75 miles. Then the scale factor is given as,

Scale factor = 5.75 / 2

Scale factor = 2.875 miles per inch

Then the actual distance from the library to the park is given as,

D = 1 x 2.875

D = 2.875 miles

The scale factor is 2.875 miles per inch. Then the actual distance from the library to the park will be 2.875 miles.

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The missing diagram is given below.

Answer:

2.875 miles per inch

Step-by-step explanation:

Answer:   C

Step-by-step explanation: We see that e is 6 for $2.70 and d is 12 for $6.00.  Half of 6 is 3 and 2.70 is less than 3 so the cheapest price can't be d. Then, we have 9 for 3.60. 3.60 divided by 9 is 0.40 per which is cheaper than a (0.42 per) so a and d can't be the cheapest. 3 for 1.38 means 0.46 per which is more expensive than a so it also cannot be b. Now we only have c and e left. We divide 2.70 by 6 and find out that the answer is 0.45 which is also more expensive than a. So, the answer is c.

Answer:

B

Step-by-step explanation:

because i said so

Step-by-step explanation:

let's say one integer is X ....

so , Another one is (X -4).....

so , X × (X - 4) = 45

x² - 4x - 45 = 0

x² - 9x + 5x - 45 = 0

x(x - 9) +5(x - 9) = 0

(x-9) × (X+5) = 0

x-9 = 0 OR x+5 = 0

X= 9 X= -5

X is not a negative integer ......

so , X = 9

X-4 = 5 .......

Answer:

44°

Step-by-step explanation:

[tex]\frac{134-46}{2}=44[/tex]

.

Solving the quadratic equation we conclude that:

The height of the rocket is defined by the quadratic equation:

[tex]d = 90t - 5t^2[/tex]

The maximum height is what we get in the vertex of the quadratic equation, such that for this quadratic equation, the vertex is at:

[tex]t = -90/(2*-5) = 9[/tex]

So the maximum height is what we get when we evaluate in t = 9:

[tex]d = 90*9 - 5*9^2 = 45[/tex]

The maximum height is 45ft.

Now we need to solve the equation for the largest value of t:

[tex]d = 90*t - 5t^2 = 0[/tex]

Rewriting it, we get:

[tex]0 = 90*t - 5*t^2\\\\0 = t*(90 - t*5)[/tex]

The maximum solution is what we get when:

0 = 90 - t*5

t = 90/5 = 18

The rocket is 18 seconds in the air.

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Answer:

25

Step-by-step explanation:

Cancel out 19^0, then exponentiate and simplify by multiplying exponents. Then simplify the terms. This leaves you with 2^-16*5^10*5^-8*2^16 which equals 5^2 which is 25

Answer:

4x=90

X=90÷4

X=22.5

5x-27=90

5x=90+27

5x=117

X=117÷5

X=23.4

So 23.4 is a larger angle

Answer:

[tex]\Large\boxed{1)~6x+7}[/tex]

[tex]\Large\boxed{2)~-7x+5y+8}[/tex]

Step-by-step explanation:

Given expression

4x + 7 + 2x

Move like terms together

= 4x + 2x + 7

= (4x + 2x) + 7

Combine like terms

[tex]\Large\boxed{=6x+7}[/tex]

Given expression

-4x + 3y - 3x + 8 + 2y

Move like terms together

= -4x - 3x + 3y + 2y + 8

= (-4x - 3x) + (3y + 2y) + 8

Combine like terms

[tex]\Large\boxed{=-7x+5y+8}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

Step-by-step explanation:

X-INTERCEPT

Plug y=0 into the equation and solve the resulting equation −6x=−7 for x.

The x-intercept:

[tex]\left(\frac{7}{6},0\right)\approx \left(1.16666666666667,0\right)[/tex]

Y-INTERCEPT

Plug x=0 into the equation and solve the resulting equation 3y=−7 for y.

The y-intercept:

[tex]\left(0, - \frac{7}{3}\right)\approx \left(0,-2.33333333333333\right)[/tex]

Answer:

[tex]x[/tex]-intercept = ([tex]-\frac{7}{6}[/tex]  , 0)

[tex]y[/tex]-intercept = (0 , [tex]-\frac{7}{3}[/tex])

Step-by-step explanation:

[tex]6x + 3y = -7[/tex]

• The x-intercept is the point at which the line crosses the x-axis, that is, where y = 0.

∴ [tex]6x + 3(0) = -7[/tex]

⇒ [tex]6x = -7[/tex]

⇒ [tex]x = \bf -\frac{7}{6}[/tex]

∴ The x-intercept is at the point ([tex]-\frac{7}{6}[/tex] , 0).

• Similarly, the y-intercept is the point at which the line crosses the y-axis, that is, where x = 0.

∴ [tex]6(0) + 3y = -7[/tex]

⇒ [tex]3y = -7[/tex]

⇒ [tex]y = \bf - \frac{7}{3}[/tex]

∴ The y-intercept is at the point (0 , [tex]-\frac{7}{3}[/tex]).

The average value of the given function is [tex]f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}[/tex].

According to the statement

we have given that the function f(x) and we have to find the average value of that function.

So, For this purpose, we know that the

The given function f(x) is

[tex]f(x) = -x^{2} + 2x +1[/tex]

And now integrate this function with the limit 0 to a then

[tex]f_{avg} = \frac{1}{b - a} \int\limits^a_0 {f(x)} \, dx = -x^{2} + 2x +1[/tex]

Now integrate this then

[tex]f_{avg} = \frac{1}{a} \int\limits^a_0 {-x^{2} + 2x +1} \, dx[/tex]

Then the value becomes according to the integration rules is:

[tex]f_{avg} = \frac{1}{a} \int\limits^a_0 {-\frac{x^{3} }{3} + \frac{2x^{2} }{2} +x} \,[/tex]

Now put the limits then answer will become as output is:

[tex]f_{avg} = \frac{1}{a} [ {-\frac{a^{3} }{3} + \frac{2a^{2} }{2} +a} \,][/tex]

Now solve this equation then

[tex]f_{avg} = [ {-\frac{a^{2} }{3} + \frac{2a }{2} +1} \,][/tex]

Now

[tex]f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}[/tex]

This is the value which represent the average of the given function in the statement.

So, The average value of the given function is [tex]f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}[/tex].

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Answer:

By definition, angles A and 1 are corresponding angles and angles B and 1 are consecutive angles. By the corresponding angles postulate, angles A and 1 are congruent, and by the consecutive angles theorem, angles B and 1 are supplementary. By the definition of supplementary angles, measures of angle B and 1 add up to 180 degrees (m<B + m<1 = 180). By definition of congruent angles, angles A and 1 have same measurement (m<A = m<1). By substitution property of equality, measures of angles A and B add up to 180 degrees (m<A + m<B = 180). By definition of supplementary angles, angles A and B are supplementary.