On the day after my birthday this year, I can truthfully say: "The day after tomorrow is Monday". On which day is my birthday?

The geometric sequence when the first term is 1/5 and the common ratio is 5 is; f(n) = f(n-1) . 5.

The geometric sequence described in the task content is one whose first term is; f(1) = 1/5.

Additionally, it follows from convention that the recursive function for a geometric sequence is; a product of a previous term and the common ratio.

Hence, we have; f(n) = f(n-1) . 5.

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Answer:

○ [tex]x = -7[/tex]

Step-by-step explanation:

Given the equation:

[tex]10(x + 10) - 4 = 11 - 5(2x + 11)[/tex],

to solve for [tex]x[/tex], we have to rearrange the equation to make [tex]x[/tex] its subject.

[tex]10(x + 10) - 4 = 11 - 5(2x + 11)[/tex]

⇒ [tex]10x + 100 - 4 = 11 - 10x -55[/tex]      [expand brackets]

⇒ [tex]10x + 96 = -44 - 10x[/tex]

⇒ [tex]10x + 10x + 96 = -44[/tex]                   [add [tex]10x[/tex] to both sides]

⇒ [tex]20x + 96 = -44[/tex]

⇒ [tex]20x = -44 - 96[/tex]                             [subtract 96 from both sides]

⇒ [tex]20x = -140[/tex]

⇒ [tex]x = \frac{-140}{20}[/tex]                                        [divide both sides by 20]

⇒ [tex]x = \bf-7[/tex]

Answer:

B) x = -7

Step-by-step explanation:

Given equation: [tex]10(x+10)-4=11-5(2x+11)[/tex]

Step 1: Distribute [tex]10[/tex] and [tex]-5[/tex] through the parentheses.

[tex]\implies 10(x)+10(10)-4=11-5(2x)-5(11)[/tex]

[tex]\implies 10x+100-4=11-10x-55[/tex]

Step 2: Simplify (combine like terms).

[tex]\implies 10x+100-4=11-10x-55[/tex]

[tex]\implies 10x+96=-44-10x[/tex]

Step 3: Add [tex]10x[/tex] to both sides.

[tex]\implies 10x+10x+96=-44-10x+10x[/tex]

[tex]\implies 20x+96=-44[/tex]

Step 4: Subtract [tex]96[/tex] from both sides.

[tex]\implies 20x+96-96=-44-96[/tex]

[tex]\implies 20x=-140[/tex]

Step 5: Divide both sides by [tex]20[/tex] to isolate [tex]x[/tex].

[tex]\implies \dfrac{20x}{20}=\dfrac{-140}{20}[/tex]

[tex]\implies \boxed{x=-7}[/tex]

Answer:

91.5[tex]cm^{2}[/tex]

Step-by-step explanation:

The area of the rectangle:

a = lw

a =(21)(3) = 63

The area of the square:

a = lw

a = (5)(5) = 25

The area of a half circle:

a=[tex]\frac{\pi r^{2} }{2}[/tex]

a= [tex]\frac{3.14(1.5^{2}) }{2}[/tex]

a = [tex]\frac{3.14(2.25)}{2}[/tex]

a = [tex]\frac{7.065}{2}[/tex]

a = 3.5 rounded to the nearest tenth

63 + 25 + 3.5 =91.5

The theorems or postulates for the given pair of angles are as follows:

2. ∠2 ≅ ∠8 → Alternate exterior angles are congruent;

3. ∠2 ≅ ∠4 → Vertically opposite angles are congruent;

4. ∠3 ≅ ∠5 → Alternate interior angles are congruent;

5. ∠3 is supplementary to ∠6 → Consecutive interior angles are supplementary;

6. ∠4 ≅ ∠8 → Corresponding angles are congruent;

Consider two lines m and n are parallel. A transversal t is intersecting the lines m and n.

So, it forms 8 angles with the lines m and n. They are ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7, and ∠8.

Based on their position, they are paired into different categories. Such as:

Interior angles: ∠3, ∠4, ∠5, ∠6

Exterior angles: ∠1, ∠2, ∠7, ∠8

Classifying the given pair of angles and their corresponding theorems:

2. ∠2 ≅ ∠8 → These angles belong to pair of Alternate exterior angles.

Theorem - "The alternate exterior angles are congruent"

3. ∠2 ≅ ∠4 → These belong to pair of vertically opposite angles.

Theorem - "The verticle angles are congruent"

4. ∠3 ≅ ∠5 → These belong to pair of alternate interior angles.

Theorem - "The alternate interior angles are congruent"

5. ∠3 is supplementary to ∠6 → These angles belong to pair of consecutive interior angles. Thus, they are supplementary.

Theorem - " The supplementary angles add up to 180°"

6. ∠4 ≅ ∠8 → These angles belong to pair of corresponding angles.

Theorem - " The corresponding angles are congruent".

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Answer:

m<C = 12.0 °

Step-by-step explanation:

Since we know that this is a non-right triangle, we must use either the law of sines or law of cosines to find m<C.  

The information we now have only allows us to use the law of sines, which focuses on proportions:

[tex]\frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}[/tex]

Thus, we can use:

[tex]\frac{sin(98)}{19}=\frac{sinC}{4}\\ 4*sin(98)=19*sinC\\ \frac{4*sin(98)}{19}=sinC\\ sin^-1\frac{4*sin98}{19}=12.033=12.0[/tex]

The sin^-1 represents the sin inverse, which you must use to find angle measures.

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

[tex]\mathsf{Formula: a^2 + b^2 = c^2}[/tex]

[tex]\textsf{Solving:}[/tex]

[tex]\mathsf{2^2 + 6^2 = c^2}[/tex]

[tex]\mathsf{2 \times 2 + 6 \times6 = c^2}[/tex]

[tex]\mathsf{4 + 36 = c^2}[/tex]

[tex]\mathsf{40 = c^2}[/tex]

[tex]\large\textsf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\frak{\sqrt{40}}}\huge\checkmark[/tex]

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

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

Answer:

Step-by-step explanation:

PS and QW intersect at U. PST = 136° and Q₁ =100°

Using the Fundamental Counting Theorem, it is found that the customer can put together 288 different outfits.

It is a theorem that states that if there are n things, each with [tex]n_1, n_2, \cdots, n_n[/tex] ways to be done, each thing independent of the other, the number of ways they can be done is:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

Bryan sells 3 types of shirts, 6 types of skirts, 8 types of bracelets and 2 types of hats, hence the parameters are given as follows:

[tex]n_1 = 3, n_2 = 6, n_3 = 8, n_4 = 2[/tex]

Hence the number of different outfits is given as follows:

N = 3 x 6 x 8 x 2 = 288

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tried my best to show the steps!! all i did was isolate to find the missing widths attached to the respective lengths (divided surface area by width) and then multiply both sides by two and add them together to find the perimeter.

The dimensions of the garden that would require the least amount of fencing are 6 ft in length and 6 ft in width.

The dimensions are the sides of a geometric shape.

The lengths of the three gardens are given as 6 ft, 9 ft, and 12 ft respectively.

If the area of the garden is 36 feet squared, then the widths for the three lengths will be as follows:

The width of the garden with a length of 6 ft [tex]= \dfrac{36}{6} = 6\ ft[/tex]

The width of the garden with a length of 9 ft [tex]= \dfrac{36}{9} = 4\ ft[/tex]

The width of the garden with a length of 6 ft [tex]= \dfrac{36}{12} = 3\ ft[/tex]

The parameters are calculated as:

The parameter of the garden with dimensions (6 ft, 6 ft) = 2(6 + 6) = 24 ft

The parameter of the garden with dimensions (9 ft, 4 ft) = 2(9 + 4) = 26 ft

The parameter of the garden with dimensions (12 ft, 3 ft) = 2(12 + 3) = 30 ft

The least of the parameter is of the garden with dimensions (6 ft, 6 ft).

Thus, the garden with dimensions (6 ft, 6 ft) would require the least amount of fencing.

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Answer:

p(t) = 2950^3t

Step-by-step explanation:

I’m not sure if this is exactly what you wanted or not. Please let me know more info and I’ll write any more answers for this question in the comments. Have a great day!!

we verified the intermidiate value theorem applies to the function f(x) = x^2 + 7x  + 1 . And the value of c is 2.

According to the given question.

We have a function.

f(x) = x^2 + 7x  + 1

As, we know that "the Intermediate Value Theorem (IVT) states that if f is a continuous function on [a,b] and f(a)<M<f(b), there exists some c∈[a,b] such that f(c)=M".

Now, we will apply the theorem for the given function f(x).

So,

f(0) = 0^2 +7(0) + 1 = 1

And,

f(9)=9² + 7(9) + 1 = 81 + 63 + 1 = 145

Here,  f(0) = 1< 19< 145 = f(9).

So, f is continous since it is a polynomial. Then the IVT applies, and such c exists.

To find, c,

We have to solve the quadratic equation f(c) =19.

This equation is

c² + 7c + 1 = 19.

Rearranging, c²+ 7c - 18=0.

Factor the expression to get

c² + 9c - 2c -18 = 0

⇒ c(c + 9) - 2( c + 9) = 0

⇒ (c - 2)(c + 9) = 0

⇒ c = 2 or -9

c = -9 is not possible beacuse it is not in the interval [0, 9].

So, the value of c is 2.

⇒ f(2) = 2^2 + 7(2) + 1 = 4 + 14 + 1 = 19

Hence, we verified the intermidiate value theorem applies to the function f(x) = x^2 + 7x  + 1 . And the value of c is 2.

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Answer:

Option 3

Step-by-step explanation:

Since the denominators are the same, you can just subtract the numerators.

[tex]\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}[/tex]

Given ,

[tex]2x = 100[/tex]

To find ,

value of x

Now ,

[tex]\longrightarrow{2x = 20}[/tex]

Dividing both sides by 2 , we get

[tex] \frac{2x}{2} = \frac{100}{2} \\ \\ \longrightarrow\boxed{ \: x = 50} [/tex]

nikal -,- xD

Answer:

Step-by-step explanation:

[tex]\bf 2x=100[/tex]

Divide both sides by 2:-

[tex]\bf \cfrac{2x}{2}=\cfrac{100}{2}[/tex]

Simplify:-

[tex]\bf x=50[/tex]

___________________

The correct option is option (d) 8 hours.

Amelia need to work for 8 hours to buy new sweater $40.

When two expressions are connected with the equals sign (=) in a math equation, it expresses the equality of the two expressions. A number that may be entered for the variable to produce a true number statement is the solution to an equation.

Let the hours Amelia need to work=h

Amalia paid=$20

Payment of each hour=$2.50

So, we can make an equation with given problem

40=20+2.50h

40-20=2.5h

20=2.5h

[tex]h=\frac{20}{2.5}[/tex]

h= 8

So, Amelia need to work for 8 hours to get $40.

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Answer:

  67 square units

Step-by-step explanation:

The area using the left-hand sum is the sum of products of the function value at the left side of the interval and the width of the interval.

The attachment shows a table of the x-value at the left side of each interval, and the corresponding function value there. The interval width is 1 unit in every case, so the desired area is simply the sum of the function values.

The approximate area is 67 square units.

Split up the interval [0, 6] into 6 equally spaced subintervals of length [tex]\Delta x = \frac{6-0}6 = 1[/tex]. So we have the partition

[0, 1] U [1, 2] U [2, 3] U [3, 4] U [4, 5] U [5, 6]

where the left endpoint of the [tex]i[/tex]-th interval is

[tex]\ell_i = i - 1[/tex]

with [tex]i\in\{1,2,3,4,5,6\}[/tex].

The area under [tex]f(x)=x^2+2[/tex] on the interval [0, 6] is then given by the definite integral and approximated by the Riemann sum,

[tex]\displaystyle \int_0^6 f(x) \, dx \approx \sum_{i=1}^6 f(\ell_i) \Delta x \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg((i-1)^2 + 2\bigg) \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg(i^2 - 2i + 3\bigg) \\\\ ~~~~~~~~ = \frac{6\cdot7\cdot13}6 - 6\cdot7 + 3\cdot6 = \boxed{67}[/tex]

where we use the well-known sums,

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

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

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

The distance from Humood's house to the school is 500m

The given parameters are:

Humood's house is (-5,7)

The school is (3,1)

The distance between both points is calculated using

[tex]d = \sqrt{(x_2- x_1)^2 + (y_2 - y_1)^2[/tex]

Substitute the known values in the above equation

[tex]d = \sqrt{(-5 - 3)^2 + (7 - 1)^2[/tex]

Evaluate

d = 10

Each unit in the graph is 50m.

So, we have

Distance =10 * 50m

Evaluate the product

Distance = 500m

Hence, the distance from Humood's house to the school is 500m

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The intersection of the sets has the following properties: Commutative law – A ∩ B = B∩ A. Associative law – (A ∩ B)∩ C = A ∩ (B∩ C) φ ∩ A = φ

do mark brainliest

The statement that describes the quotient of 3 over 10 division sign2 over 5 is A. The maximum amount of oil the bowl can hold is Fraction 3 over 4 cup.

From the information given, we are told that a bowl holds fraction 3 over 10 cups of oil when it is Fraction 2 over 5 full.

Therefore, the statement that best describes the quotient of 3 over 10 division sign2 over 5 will be that the maximum amount of oil the bowl can hold is fraction 3 over 4 cup.

In conclusion, the correct option is A.

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[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

Let's calculate its discriminant ~

[tex]\qquad \sf  \dashrightarrow \: {t}^{2} + \cfrac{17}{2} t - 5 = 0[/tex]

[ Multiply both sides by 2 ]

[tex]\qquad \sf  \dashrightarrow \: 2 {t}^{2} + 17t - 10[/tex]

[tex]\qquad \sf  \dashrightarrow \: discriminant = {b}^{2} - 4ac[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = (17) {}^{2} - (4 \times 2 \times - 10)[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 289 - ( - 80)[/tex]

[tex]\qquad \sf  \dashrightarrow \: d = 369[/tex]

[tex]\qquad \sf  \dashrightarrow \: \sqrt {d }= 3 \sqrt{41} \approx19.209 [/tex]

So, by quadratic formula :

[tex]\qquad \sf  \dashrightarrow \: t = \dfrac{ - {b}^{} \pm \sqrt{d} }{2a} [/tex]

[tex]\qquad \sf  \dashrightarrow \: t = \dfrac{ - {17}^{} \pm \sqrt{369} }{2 \times 2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \:t = \cfrac{ - 17 - 19.209}{4} \: \: and \: \: t = \dfrac{-17+19.209}{4} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \:t = \cfrac{ - 36.209}{4} \: \: and \: \: t = \dfrac{2.209}{4} [/tex]

[tex]\qquad \sf  \therefore \: t = - 9.052 \: \: \: or \: \: \: t = 0.552[/tex]

The probability that the first interview occurs on the fifth resume that you sent is 0.2592.

According to the given question.

The probability of being called for an interview, p = 0.40.

So, the probability of not being called for an interview, q = 1 - 0.40 = 0.60

As we know that binomial distribution summarizes the number of trials, or observations when each trial has the same probability of attaining one particular value. The binomial distribution determines the probability of observing a specified number of successful outcomes in a specified number of trials.

Therefore, the probability that the first interview occurs on the fifth resume that you send out

= [tex]^{5} C_{1} (0.40)^{1} (0.60)^{4}[/tex]

= 5(0.40)(0.1296)

= 0.2592

Hence, the probability that the first interview occurs on the fifth resume that you sent is 0.2592.

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Answer:

[tex]y = - \frac{1}{4} x - 10[/tex]

Step-by-step explanation:

The negative reciprocal of 4 is -1/4.

Let's substitute in our values, to find the y-intercept:

[tex]y = - \frac{1}{4} x + c[/tex]

[tex] - 11 = - 1 + c[/tex]

[tex]c = - 10[/tex]

Finally, our full equation is:

[tex]y = - \frac{1}{4} x - 10[/tex]

Answer:

[tex]y=-\frac{1}{3}x+7[/tex]

Step-by-step explanation:

Parallel lines have the same slope, so the slope of the line we need to find is -1/3.

Substituting into point-slope form and converting to slope-intercept form,

[tex]y-5=-\frac{1}{3}(x-6) \\ \\ y-5=-\frac{1}{3}x+2 \\ \\ y=-\frac{1}{3}x+7[/tex]

The number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

Simultaneous equation is an equation which involves the solving for two unknown values at the same time.

x + y = 3,500

x – y = 2,342

Add both be equation

x + x = 3,500 + 2,342

2x = 5,842

x = 5,842 ÷ 2

x = 2,921

Substitute x = 2,921 into

x – y = 2,342

2,921 - y = 2, 342

-y = 2,342 - 2,921

-y = -579

y = 579

Therefore, the number of contemporary titles and classic titles in Jarred DVDs collection is 2,921 and 579 respectively.

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The two ratios in the tables shown which have a common denominator you could use to compare is; 3/12 and 4/12

Machael's ratio:

lemons to water = 1:4

Equivalent ratio

= 3 : 12

= 4 : 16

Sondra's ratio:

lemons to water = 2 : 6

Equivalent ratio

= 4 : 12

= 6 : 18

Therefore, the two ratios in the tables shown which have a common denominator you could use to compare is; StartFraction 3 Over 12 EndFraction and StartFraction 4 Over 12 EndFraction

Complete question

Michael and Sondra are mixing lemonade. In Michael’s lemonade, the ratio of lemons to water is 1:4. In Sondra’s lemonade, the ratio of lemons to water is 2:6. Several equivalent ratios for each mixture are shown in the ratio tables. Imagine that you want to compare Michael’s ratio to Sondra’s ratio. Which two ratios in the tables shown have a common denominator you could use to compare?

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There is a maximum value of 7/6 located at (x, y) = (5/6, 7).

The function given to us is f(x, y) = xy.

The constraint given to us is 6x + y = 10.

Rearranging the constraint, we get:

6x + y = 10,

or, y = 10 - 6x.

Substituting this in the function, we get:

f(x, y) = xy,

or, f(x) = x(10 - 6x) = 10x - 6x².

To find the extremum, we differentiate this, with respect to x, and equate that to 0.

f'(x) = 10 - 12x ... (i)

Equating to 0, we get:

10 - 12x = 0,

or, 12x = 10,

or, x = 5/6.

Differentiating (i), with respect to x again, we get:

f''(x) = -12, which is less than 0, showing f(x) is maximum at x = 5/6.

The value of y, when x = 5/6 is,

y = 12 - 6x,

or, y = 12 - 6*(5/6) = 7.

The value of f(x, y) when (x, y) = (5/6, 7) is,

f(x, y) = xy,

or, f(x, y) = (5/6)*7 = 7/6.

Thus, there is a maximum value of 7/6 located at (x, y) = (5/6, 7).

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A logarithmic equation exists as an equation that uses the logarithm of an expression containing a variable. The value of the logarithmic equation x = log 2.97/log 1.13.

A logarithmic equation exists as an equation that applies the logarithm of an expression having a variable. To estimate exponential equations, first, see whether you can note both sides of the equation as powers of the same number.

Given: [tex]$1.13^x = 2.97[/tex]

Taking log on both sides, we get

log [tex]$1.13^x[/tex] = log 2.97

x log 1.13 = log 2.97

x = log 2.97/log 1.13

Therefore, the value of logarithmic equation x = log 2.97/log 1.13.

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The equation that represents the function g(x) is g(x) = log₅(x - 3)

The equation of the function f(x) is given as:

f(x) = log₅(x)

From the graph, we can see that the function g(x) is 3 units to the right of the function f(x)

This means that

g(x) = f(x - 3)

The function f(x - 3) is calculated as follows

f(x - 3) = log₅(x - 3)

Substitute f(x - 3) = log₅(x - 3) in g(x) = f(x - 3)

g(x) = log₅(x - 3)

Hence, the equation that represents the function g(x) is g(x) = log₅(x - 3)

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