Solve for x. You must show all of your work to receive credit.

Check the picture below.

Using the given definition and [tex]\Delta x=\frac{0-(-2)}n=\frac2n[/tex], we have

[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \sum_{i=1}^n \left(7\left(-2+\frac{2i}n\right)^2 + 7\left(-2+\frac{2i}n\right)\right) \frac2n \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac2n \sum_{i=1}^n \left(14 - \frac{42i}n + \frac{28i^2}{n^2}\right)[/tex]

Recall the well-known power sum formulas,

[tex]\displaystyle \sum_{i=1}^n 1 = \underbrace{1 + 1 + 1 + \cdots + 1}_{n\,\rm times} = n[/tex]

[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]

[tex]\displaystyle \sum_{i=1}^n i^2 = 1 + 4 + 9 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6[/tex]

Reducing our sum leads to

[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \frac2n \left(\frac{7n}3 - 7 + \frac{14}{3n}\right) = \lim_{n\to\infty} \left(\frac{14}3 - \frac{14}n + \frac{28}{3n^2}\right)[/tex]

As [tex]n[/tex] goes to ∞, the rational terms containing [tex]n[/tex] will converge to 0, and the definite integral converges to

[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \boxed{\frac{14}3}[/tex]

Answer:

7

Step-by-step explanation:

→ Find the scale factor

30 ÷ 25 = 1.2

→ Multiply answer by 20

20 × 1.2 = 24

→ Equate equation to 24

4x - 4 = 24

→ Add 4 to both sides

4x = 28

→ Divide both sides by 4

x = 7

Since Hassan's estimation is between 0.4 and 0.5, hence the Hassan is incorrect because √0.15 is  less than 0.4

The square root of numbers is is expressed using the square root sign. In order to determine the square root os 0.15 given, we need need to determine the square of perfect square before and after the given number.

For the square root of √0.09

√0.09 = 0.3

Similarly for the square root of √0.16

√0.16 = 0.4

Since the resulting value is 0.3 and 0.4, hence the square root of 0.15 must be between these two values on the number line.

Since Hassan's estimation is between 0.4 and 0.5, hence the Hassan is incorrect because √0.15 is  less than 0.4

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Answer:

None of these answers are correct.

Step-by-step explanation:

[tex]QR=\frac{25+45}{2}=35[/tex]

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

By Trapezoid mid - segment theorem :

[tex] \qquad❖ \: \sf \:QR = \cfrac{MN+ OP}{2} [/tex]

[tex] \qquad❖ \: \sf \:QR= \cfrac{25+ 45}{2} [/tex]

[tex] \qquad❖ \: \sf \:QR = \cfrac{70}{2} [/tex]

[tex] \qquad❖ \: \sf \:QR= 35 \: \: units[/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

The simplified form of the given expression as a fraction is 7x/x-7

Fractions  are written as a ratio of two integers. For instance a/b is a fraction where a and b are integers.

Given the expression below;

(1/7+1/x)/(1/49+1/x²)

Find the LCM to have:

(x+7/7x)/(x²-49)/49x²

Divide to have:

x+7/7x * 49x²/(x²-49)

(x+7) * 7x/(x+7)(x-7)

Cancel out the like terms to have:

1 * 7x/x-7

= 7x/x-7

Hence the simplified form of the given expression as a fraction is 7x/x-7

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Step-by-step explanation:

volume = area of circle × hight of cylinder ....

V = πr² × h

so V = πr²h

Answer:

  10831 mm³

Step-by-step explanation:

The volume of any solid figure with a uniform cross section parallel to the base can be found using the volume formula ...

  V = Bh

where B is the area of the base, and h is the height perpendicular to the base.

The base of this cylindrical stack of pennies is a circle of diameter 19.05 mm. The area formula for a circle in terms of diameter is ...

  A = (π/4)d²

The stack of pennies has a base area of ...

  A = (π/4)(19.05 mm)² ≈ 285.023 mm²

The volume formula tells us the stack of pennies has a volume of ...

  V = Bh

  V = (285.023 mm²)(38 mm) ≈ 10830.9 mm³

The volume of the stack of pennies is approximately 10831 mm³.

Answer:

Step-by-step explanation:

(1). A = π r²

A = 4 π

Area of shaded region is [tex]\frac{4\pi *260}{360}[/tex] = [tex]\frac{26}{9} \pi[/tex] or [tex]\frac{26\pi }{9}[/tex]

(2). C = 4 π

In this case, the length of the bigger arc and the area of shaded region happen to be the same.

The length of the arc ADB is  [tex]\frac{26}{9} \pi[/tex] or [tex]\frac{26\pi }{9}[/tex]

Answer:

10(x-3) is equivalent to 10x – 30

Step-by-step explanation:

First you have to take everything of out the parentheses and then simplify if needed.  

Answer: 10(x-3) is equivalent to 10x – 30

[tex]~~~~~~ \textit{Annual Percent Yield Formula} \\\\ ~~~~~~~~~~~~ APY=\left(1+\frac{r}{n}\right)^{n}-1 ~\hfill \begin{cases} r=rate\to 10\%\to \frac{10}{100}\dotfill &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12 \end{cases} \\\\\\ APY=\left(1+\frac{0.1}{12}\right)^{12}-1\implies APY=\left( \frac{121}{120} \right)^{12}-1\implies APY\approx 0.1047[/tex]

The expression involving exponents to represent the shaded area in square inches of the diagram is: 6²- (3² + 2²). The shaded area in squares inches of the diagram is: 23 square inches.

The expression is:

6²- (3² + 2²)

The shaded area:

Shaded area=6²- (3² + 2²)

Shaded area=36-(9+4)

Shaded area=36-13

Shaded area=23 square inches

Therefore the expression involving exponents to represent the shaded area in square inches of the diagram is: 6²- (3² + 2²). The shaded area in squares inches of the diagram is: 23 square inches.

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Answer: (0,6)

Step-by-step explanation:

The solution is where the graphs intersect.

The solution to the inequality x-13<=7+4x is x >= -20/3

The inequality expression is given as:

x-13<=7+4x

Add 13 to both sides of the inequality

x <= 20 + 4x

Subtract 4x from both sides of the inequality

-3x <= 20

Divide both sides of the inequality by -3

x >= -20/3

Hence, the solution to the inequality x-13<=7+4x is x >= -20/3

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Complete question

Solve the inequality for x

x-13<=7+4x

Answer: C: 43

Step-by-step explanation:

As we are calling the function with 2 for x, we can substitute 2 for every x we see in the function and solve.

[tex]f(2)=2(2)^3-19(2)^2+28(2)+47\\=2(8)-19(4)+28(2)+47\\=16-76+56+47\\=43[/tex]

Hence, f(2) is 43.

The upper quartile for the given data is 51 .

Given data: {51,49,52,46,50,38,38,45,34,52,46}

Data in arranged form: {34,38,38,45,46,46,49,50,51,52,52}

Find the median first, then the upper quartile. There are eleven data points, thus look at term six to determine the median as there are five data points on any side. The median is 46 because the sixth term is 46.

The upper extreme, 52, and the median are then used to determine the upper quartile; make sure not to include the median data point when dividing the groups. The upper quartile is the third term (between the median and upper extreme) because there would be two data points on each side if there were only five data points from the median to the upper extreme. The third term, which is 51, falls into the upper quartile.

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Using proportions, it is found that he bikes:

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

In this problem, the proportion is that he bikes 9.5 miles each half hour, hence:

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The conclusion of the Hypothesis Conclusion is; that there is sufficient evidence to support the claim that the fund has moderate risk.

We are given;

Sample size; n = 36

Population standard deviation; σ₀ = 10

Sample standard deviation; s = 4.95

Significance level; α = 0.05

Claim: Standard deviation less than 10

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis needs to contain an equality and the value mentioned in the claim. If the claim is the null hypothesis, then the alternative hypothesis states the opposite of each other. Thus;

Null Hypothesis; H₀: σ = 10

Alternative Hypothesis; H₁: σ < 10

Compute the value of the test statistic:

χ2 = [(n - 1)/(σ²)] * s²

χ2 = [(36 - 1)/(10²)] * 4.95²

χ2 = 8.576

The critical value of the left-tailed test is given in the row with df = n - 1 = 36 - 1 = 35 and in the column with 1 − α = 0.95 of the chi-square distribution table online, we have;

χ2_{1 - α} = 21.77

The rejection region then contains all values smaller than 21.77

If the test statistic is in the rejection region, then reject the null hypothesis:

8.576 < 13.848

Thus, we will reject H₀ and conclude that there is sufficient evidence to support the claim that the fund has moderate risk.

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The required rewritten equation in terms of variable 'r' is r = -u/s²t. By applying simple arithmetic operations, the required equation is obtained.

Consider an equation c = ax + by. Solving for variable 'a'.  

Step1: Write the required variable terms on one side

Step2: Add/Subtract like terms if any or take common if any

Step3: Divide/Multiply the coefficient of the required variable

Step4: simplify the obtained terms for the required variable

As follows:

c = ax + by

⇒ ax = by - c

⇒ ax/x = (by - c)/x

∴ a = (by - c)/x

The given equation is u = -s²rt

dividing by 't' on both sides:

⇒ u/t = -s²rt/t

⇒ u/t = -s²r

dividing by 's²' on both sides:

⇒ u/s²t = -s²r/s²

⇒ u/s²t = -r

∴ r = -u/s²t

Thus, the required equation for variable 'r' is r = -u/s²t.

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Answer + Step-by-step explanation:

the correct question:

For r ≠ 1 ,Prove using the mathematical induction method that :

[tex]1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}[/tex]

………………………………………………………………………………………………………………

for n = 0 :

1⁰ = 1  and  (1 - r⁰⁺¹)/(1 - r) = (1 - r)/(1 - r) = 1

Then the property is true for n = 0.

For n ≥ 0 :

Suppose

[tex]1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}[/tex]

And prove that

[tex]1+\cdots+r^{n+1} =\frac{1-r^{n+2}}{1-r}[/tex]

Since :

[tex]1+\cdots+r^{n+1} =(1+\cdots+r^n)+r^{n+1}[/tex]

Then

[tex]1+\cdots+r^n+r^{n+1} =\frac{1-r^{n+1}}{1-r}+r^{n+1}[/tex]

[tex]= \frac{1-r^{n+1}+r^{n+1}(1-r)}{1-r}[/tex]

[tex]= \frac{1-r^{n+2}}{1-r}[/tex]

Then according to the mathematical induction method

[tex]1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}[/tex]

Where n is a natural number and r ≠ 1.

[tex]\frac{4}{7}(504)=\boxed{288}[/tex]

If the sum of the length and width of rectangle is 60 and rectangle is having maximum area then the dimensions are 30 units each.

Given that the sum of length and breadth of rectangle is 60.

We are required to find the dimensions of the rectangle that will have the maximum area. Area is basically how much part of surface is being covered by that particular shape or substance.

Let the length of rectangle be x.

According to question the breadth will be (60-x).----2

Area of rectangle=Length *Breadth

A=x(60-x)

A=60x-[tex]x^{2}[/tex]

Differentiate A with respect to x.

dA/dx=60-2x

Again differentiate with respect to x.

[tex]d^{2} A/dA^{2}[/tex]=-2x

-2x<0

So the area is maximum because x cannot be less than or equal to 0.

Put dA/dx=0

60-2x=0

60=2x

x=30

Put the value of x in 2 to get the breadth.

Breadth=60-x

=60-30

=30

Hence if the sum of the length and width of rectangle is 60 and rectangle is having maximum area then the dimensions are 30 units each.

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Answer:

10x

Step-by-step explanation:

g(x)=5x

f(x)=2x

f(g(x))=f(5x)

f(g(x))=2*5x=10x

The true statement about the series [tex]\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}[/tex] is that (a) the series diverges

The series is given as:

[tex]\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}[/tex]

Take the limit of the function to infinity

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)}[/tex]

This gives

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty * (\infty +3)}[/tex]

Evaluate the sum

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty * \infty}[/tex]

Evaluate the product

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty}[/tex]

Evaluate the quotient

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = 0[/tex]

Since the limit is 0, then it means that the series diverges

Hence, the true statement about the series [tex]\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}[/tex] is that (a) the series diverges

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Answer:

B. The series converges to [tex]\displaystyle{\frac{11}{2}}[/tex].

Step-by-step explanation:

Before evaluating the infinite series, the expression can be decomposed as the sum of two fractions (partial fraction decomposition) as follows.

Let [tex]\textit{A}[/tex] and [tex]\textit{B}[/tex] be constants such that

                                       [tex]{\displaystyle{\frac{9}{n\left(n+3\right)}}}} \ \ = \ \ \displaystyle{\frac{A}{n} \ \ \ + \ \ \frac{B}{n+3}}[/tex]

Multiply both sides of the equation by the denominator of the left fraction,

[tex]n\left(n+3\right)[/tex], yielding

                                              [tex]9 \ \ = \ \ A\left(n+3\right) \ \ + \ \ B \-\hspace{0.045cm} n[/tex]

Now, let [tex]n \ = \ 0[/tex], thus

                                             [tex]\-\hspace{0.2cm} 9 \ \ = \ \ A\left(0 + 3\right) \ + \ B\left(0\right) \\ \\ 3 \-\hspace{0.035cm} A \ = \ \ 9 \\ \\ \-\hspace{0.11cm} A \ \ = \ \ 3[/tex].

Likewise, let [tex]n \ = \ -3[/tex], then

                                            [tex]\-\hspace{0.5cm} 9 \ \ = \ \ A\left(-3 + 3\right) \ + \ B\left(-3\right) \\ \\ -3 \-\hspace{0.035cm} B \ = \ \ 9 \\ \\ \-\hspace{0.44cm} B \ \ = \ \ -3[/tex]

Hence,

                                       [tex]\displaystyle{\sum_{n=1}^{\infty} {\frac{9}{n\left(n+3\right)}}} \ = \ \displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right)[/tex].

First and foremost, write the nth partial sum (first nth terms) of the series,

          [tex]\displaystyle\sum_{n=1}^{n} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \displaytstyle{\frac{3}{1} \ - \frac{3}{4} + \ \frac{3}{2} \ - \frac{3}{5} \ + \frac{3}{3} \ - \frac{3}{6} \ + \frac{3}{4} \ - \frac{3}{7}} \\ \\ \\ \-\hspace{3.58cm} + \ \displaystyle{\frac{3}{5} \ - \ \frac{3}{8} \ + \ \frac{3}{6} \ - \ \frac{3}{9} \ + \ \frac{3}{7} \ - \ \frac{3}{10}} \\ \\ \\ \-\hspace{3.58cm} + \ \ \dots[/tex]

                                                [tex]+ \ \ \displaystyle{\frac{3}{n-3} \ - \ \frac{3}{n} \ + \ \frac{3}{n-2} \ - \ \frac{3}{n+1}} \\ \\ \\ \ + \ \frac{3}{n-1} \ - \ \frac{3}{n+2} \ + \ \frac{3}{n} - \ \frac{3}{n+3}}[/tex].

Notice that the expression forms a telescoping sum where subsequent terms cancel each other, leaving only

        [tex]\displaystyle\sum_{n=1}^{n} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \displaytstyle{\frac{3}{1} \ + \ \frac{3}{2} \ + \frac{3}{3} \ - \ \frac{3}{n+1}} - \ \frac{3}{n+2} \ - \ \frac{3}{n+3}}}[/tex].

To determine if this infinite series converges or diverges, evaluate the limit of the nth partial sum as [tex]n \ \rightarrow \ \infty[/tex],

         [tex]\displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \lim_{n \to \infty} \left(\displaytstyle{\frac{11}{2} \ - \ \frac{3}{n+1} \ - \ \frac{3}{n+2} \ + \ - \ \frac{3}{n+3}\right) \\ \\ \\ \-\hspace{3.25cm} = \ \ \ \displaystyle{\frac{11}{2} \ - \ 0 \ - \ 0 \ - \ 0} \\ \\ \\ \-\hspace{3.25cm} = \ \ \ \displaystyle{\frac{11}{2}[/tex]

Step-by-step explanation:

second one

explanation: trust me bro

The area of the trapezoid is 140 square feet

The area of a shape is the amount of space on that shape

The dimensions of the trapezoid are given as:

Height = 10 feet

Parallel bases = 8 feet and 20 feet

The area of a trapezoid is calculated using

Area = 0.5 * (Sum of parallel bases) * Height

Substitute the known values in the above equation

Area = 0.5 * (8 + 20) * 10

Evaluate the equation

Area = 140

Hence, the area of the trapezoid is 140 square feet

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Answer:

I dont know if this is right but i got -7/3.

Sorry if its wrong

Answer:

None of these

Step-by-step explanation:

[tex]\frac{360}{n}=24 \implies n=15[/tex]

This is called a pentadecagon.

Answer: well one is

Bc its the smallest besides zero, but zero is neither odd or even

Step-by-step explanation:

Answer:

125

Step-by-step explanation:

The sum of two interior angles in a triangle is equal to an exterior angle that is supplementary to the third interior angle.

We can write the following equation according to this information and that will help us find the value of x:

65 + 60 = x add like terms

125 = x is the answer we are looking for.

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textsf{equation:}[/tex]

[tex]\large\textsf{a = 60 + 65}[/tex]

[tex]\huge\textsf{solving:}[/tex]

[tex]\large\textsf{a = 60 + 65}[/tex]

[tex]\large\textsf{60 + 65 = a}[/tex]

[tex]\huge\textsf{simplify it:}[/tex]

[tex]\large\textsf{a = 125}[/tex]

[tex]\huge\textsf{therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\mathsf{a =} \frak{125}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]