A phone company offers two monthly plans. Plan A costs $20 plus an additional $0.08 for each minute of calls. Plan B costs $16 plus an additional $0.13 for each minute of calls. For what amount of calling do the two plans cost the same? minutes What is the cost when the two plans cost the same? ​

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

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

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]

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]

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:

sorry i thought i knew it

Step-by-step explanation:

Answer:

3rd option

Step-by-step explanation:

using the rules of exponents

[tex]\frac{a^{m} }{a^{n} }[/tex] = [tex]a^{(m-n)}[/tex] : nm > n

[tex]\frac{a^{m} }{a^{n} }[/tex] = [tex]\frac{1}{a^{(n-m)} }[/tex] : n > m

[tex]\frac{6a^2b^{-2} }{8a^{-3b^3} }[/tex] ← separate the variables

= [tex]\frac{6}{8}[/tex] × [tex]\frac{a^2}{a^{-3} }[/tex] × [tex]\frac{b^{-2} }{b^3}[/tex]

= [tex]\frac{3}{4}[/tex] × [tex]a^{2-(-3)}[/tex] × [tex]\frac{1}{b^{3-(-2)} }[/tex]

= [tex]\frac{3}{4}[/tex] × [tex]a^{2+3}[/tex] × [tex]\frac{1}{b^{3+2} }[/tex]

= [tex]\frac{3}{4}[/tex] × [tex]a^{5}[/tex] × [tex]\frac{1}{b^{5} }[/tex]

= [tex]\frac{3a^{5} }{4b^{5} }[/tex]

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.

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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We cannot get further information about the dimensions of the piece since the number of variables is greater than the number of equations.

By geometry we know that the area of the piece of metal is equal to the product of its length and width, then we must find two real numbers such that:

l · w = 27.75, where l, w > 0.

Unfortunately, we cannot get further information about the dimensions of the piece since the number of variables is greater than the number of equations. We need at least one equation to find an unique solution.

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The linear equation that represents the relationship between earnings and tips will be y = 0.75x + 50.

The linear equation is given as,

y = mx + c

Where m is the slope of the line and c is the y-intercept of the line.

Working as a waitress is Amanda. She receives $50 every day in addition to 75% of the tips her clients leave.

Let 'x' be the tip amount and 'y' be the total earning.

y = 0.75x + 50

Check:

At (20,65), then we have

65 = 0.75(20)+50

65 = 15+50

65=65

At (50,87.5), then we have

87.5 = 0.75(50)+50

87.5 = 37.5+50

87.5=87.5

At (100,125), then we have

125 = 0.75(100)+50

125 = 75+50

125=125

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Check the picture below.

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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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:

None of these

Step-by-step explanation:

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

This is called a pentadecagon.

Check the picture below.

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

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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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]

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.

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:  B(2, -3) and C(-2, 3) which is choice 3

Further Explanation:

The rule for a y-axis reflection is [tex](\text{x}, \text{y}) \to (-\text{x}, \text{y})[/tex] meaning the x coordinate flips from positive to negative, or vice versa. The y coordinate stays the same.

Therefore, point A(-2,-3) moves to B(2, -3) when reflecting over the y-axis.

A similar rule is [tex](\text{x}, \text{y}) \to (\text{x}, -\text{y})[/tex] to describe an x-axis reflection. This time the y coordinate flips in sign, and the x coordinate stays the same.

In this case, we move from A(-2,-3) to C(-2, 3)

The graph of all three points is shown below. I used GeoGebra to make the graph, but Desmos is another option. I also recommend graphing by hand on graph paper to get practice that way if possible.

Answer:

Step-by-step explanation:

y - 4 = 5/4(x - 8)

y - 4 = 5/4x - 10

y = 5/4x - 6

Answer: (0,6)

Step-by-step explanation:

The solution is where the graphs intersect.

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:

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

Sorry if its wrong

Answer:

10x

Step-by-step explanation:

g(x)=5x

f(x)=2x

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

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

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

[tex]x + y = 10\implies y = -x+10\implies y=\stackrel{\stackrel{m}{\downarrow }}{-1}x+10 \leftarrow \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]

so, we're really looking for the equation of al ine whose slope is -1 and that passes through (-1 , 5)

[tex](\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\hspace{10em} \stackrel{slope}{m} ~=~ -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-1}( ~~ x-\stackrel{x_1}{(-1) ~~ }) \\\\\\ y-5 = -(x+1)\implies y-5=-x-1\implies y=-x+4[/tex]