Last month, jennifer earned $180 for her after-school job. if she was paid $7.50 per hour, how many hours did jennifer work?

The equation of the circle that passes through the point (0 , 4) and has a center at the origin is x^2 + y^2 = 16.

Using the distance formula, get the radius of the circle by solving for the distance between the center and the point (0 , 4).

radius = distance = √(x2 - x1)^2 + (y2 - y1)^2

radius = √(0 - 0)^2 + (4 - 0)^2

radius= √0 + 16

radius = 4

The standard form of the equation of the circle is given by (x - h)^2 + (y - k)^2 = r^2, where (h , k) is the location of the center and r is the radius of the circle.

Given the radius and center of the circle, substitute these values to the standard form of the equation of the circle.

(x - h)^2 + (y - k)^2 = r^2

where (h , k) = (0 , 0)

r = 4

(x - 0)^2 + (y - 0)^2 = 4^2

x^2 + y^2 = 16

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The slope is 23, from x>=1 , the domain is (1,∞) and the range is {f: A→B; f(x) = 23x}.

Given,

Cost for first day = $39

Cost for each additional day = $23

Let x denote the number if days

Y denote the total cost

(a) As the cost for the first day to rent a zoom $39, so its domain is from (x=0 to x=1)

i.e., (0,1] and Range is $39 --> for the first day

For each additional day, the cost increases by $ 23.

Therefore, slope is 23, from x>=1.

(b) For the first day = $39

Therefore, cost is fixed for the first day its domain is (0,1] and Range is $39

For each additional day, cost increases by $ 23

Therefore, slope after x=1 is 23

Domain = (1,∞)

Ranges = {f: A→B; f(x) = 23x}

Ranges will be set of positive integers of multiple of 23

i.e., {23,46,69…….}

(c) Set of equations to model the relationship.

Hence the equation is,

[tex]\mathrm{y}=\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c}\$ 39,0 \leq x \leq 1 \\39+23 x, x > 1\end{array}\right.[/tex].

Hence, the slope is 23, from x>=1 , the domain is (1,∞) and the range is {f: A→B; f(x) = 23x} can be used to model relationship between the price of renting a car and the number of days you rent it.

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Hello,

[tex]2(x - 3) + 2x = 22[/tex]

[tex]2x - 6 + 2x = 22[/tex]

[tex]4x - 6 = 22[/tex]

[tex]4x = 22 + 6[/tex]

[tex]4x = 28[/tex]

[tex]x = \frac{28}{4} = 7[/tex]

Answer:

f = - [tex]\frac{g}{x-1}[/tex]

Step-by-step explanation:

fx = f - g ( subtract f from both sides )

fx - f = - g ( factor out f from each term on the left side )

f(x - 1) = - g ( divide both sides by (x - 1) )

f = - [tex]\frac{g}{x-1}[/tex]

The tangent to the given curve at the point (1,0) is: y=6x-6.

The first derivative f'(x), the slope (tangent) of the original curve, may be obtained at every point (x,y) on the curve because the supplied curve is a continuous function, making Eq. 1 differetiable at every x.

The slope of the tangent at x=1 is given by the first derivative of Eq. 1: f'(1)=9(1)2-3=6.

A line that passes through the point (x1, y1) and has a slope of m has the equation:

(y-y1)=m(x-x1) ··· Eq.2 (the point- slope form of linear function)

Here, x1=1, y1=0, and m=f'(x1)=6. Add these numbers to Eq. 2. It is possible to repeat and simplify Eq. 2 as follows: y-0=6(x-1) ⇒ y=6x-6

Consequently, y=6x-6 is the tangent to the given curve at position (1, 0).

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In other words, the triangles are consistent if the hypotenuse and one of the legs of one right triangle agree with the hypotenuse and one of the legs of another right triangle.

Congruent triangles are two triangles that are the same size and shape. Two congruent triangles remain congruent even if we flip, turn, or rotate one of them. Two triangles must have the same angles and, as a result, must be congruent if their sides are the same.

Two right triangles are supposed to be consistent on the off chance that they are of a similar shape and size. All in all, two right triangles are supposed to be consistent in the event that the proportion of the length of their comparing sides and their relating points is equivalent.

Right triangles are harmonious in the event that both the hypotenuse and one leg are a similar length. These triangles are harmonious by HL or hypotenuse leg.

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The value of h is (3fg +5i²) / 3 using the concept of algebraic operations.

Any of the common arithmetic operations, such as addition, subtraction, multiplication, division, raising to a whole number power, and taking roots, are considered basic algebraic operations in mathematics (fractional power).

Given:

fg= h – 5/3 × i²

Add 5/3i² both sides

h= fg+ 5/3i²

h= (3fg +5i²) / 3

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The value of h in the given operation fg= h – 5/3 × i² is (3fg +5i²) / 3.

So, fg= h – 5/3 × i².

Addition of 5/3i² on both sides as follows:

Therefore, the value of h in the given operation is (3fg +5i²) / 3.

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3 . (4-8) / 2

= 3. (-4) /2

= 3. ( -2)

=  -6

In algebraic equation or polynomial equation is an equation of the frame P=0 where P could be a polynomial with coefficients in a few field, regularly the field of the rational numbers.

algebraic equation, statement of the equality of two expressions defined by applying to a set of factors the algebraic operations, namely, expansion, subtraction, increase, division, raising to a control, and extraction of a root.

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A glide reflection is a translation followed by a reflection in a line parallel to the translation vector.

A glide translation is what?

when a figure is subjected to one transformation before another is applied?

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The  probability that a randomly chosen dog is a female or over 5 years old is 0.68 .

 Age                       Male             Female  

Under 1 year            6                        5

1 to5 year                  8                         7

6 to10 years              4                         6

Over 10 years            3                         5

From the table, we can estimate that there are 23  f_emale dogs, 18 are over 5 years old,  and there are 11 f_emale dogs who are over 5 years old.

Therefore,  the total number of dogs which are f_emale and over 5 years old =  23 + 18 – 11 = 30

This means there total of 44 dogs in total.

A be the event the  randomly chosen dog is f_emale and  over 5 years old, The probability of the event will be, P(A)=  30/44 = 0.68.

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Answer: y=31-5x

Step-by-step explanation:

subtract the 5x to the other side to get y.

The radius of a circle with an area of π/4 square units is B. 0.5 units

Area is the quantity that expresses the extent of a region on the plane or on a curved surface. The area of a plane region or plane area refers to the area of a shape or planar lamina, while surface area refers to the area of an open surface or the boundary of the three-dimensional object.

Now it is given that,

Area of circle =  π/4 square units

Since we know that,

Area of circle =  πr²

where r is the radius of circle.

⇒ r² =  Area of circle / π

Put the values we get,

⇒ r² = (π/4)/ π

⇒ r² = 1/4

Taking square root both the side we get,

⇒ r = 1/2 units

or,  r = 0.5 units

Thus, the radius of a circle with an area of π/4 square units is B. 0.5 units

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The exterior angle is acute, the sum of the farthest interior angles must also be acute, implying that the third angle is obtuse. As a result, every triangle should be obtuse.

An obtuse-angled triangle has one of its interior provides a clear view than 90°.. If one angle in an obtuse triangle measures upwards of 90°, the combination of the remaining two angles is a little less than 90°.

if the exterior angle is acute, this same sum of something like the remote interior angles must also be acute, implying that the third angle is obtuse. As a result, the triangle must be obtuse. Furthermore, because the exterior angle and A form a linear pair, A must be obtuse because consecutive acute angles cannot form a linear pair.

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Answer:

Could you explain the problem a bit more?

Step-by-step explanation:

The equation of the line in point-slope form, which passes through the pair of points located at (0 , 1/2) and (5/7 , 0), is y - 1/2 = -7/10 x

The equation of a line can be expressed in three different forms: standard form, slope-intercept form, and point-slope form.

The standard form of an equation of a line is expressed as ax + by = c, where a and b, if all possible must be integers, are the coefficients of variable x and y, respectively, and c is a constant. Meanwhile, slope-intercept form is given by the formula y = mx + b, where m is the slope of the line and b is the y- intercept. On the other hand, given the slope m and a point on the line (x , y), we can express the equation in point-slope form, (y - y1) = m(x - x1).

Getting the slope of the two points: (0 , 1/2) and (5/7 , 0).

m = (y2 - y1)/(x2 - x1)

m = (0 - 1/2)/(5/7 - 0)

m = (-1/2)/(5/7)

m = -7/10

Using the point slope form, plug in the values of the slope and one point to set up the equation.

(y - y1) = m(x - x1)

where m = -7/10 and (x1 , y1) = (0 , 1/2)

(y - 1/2) = -7/10(x - 0)

y - 1/2 = -7/10 x

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The exact value of the expression is -1/2

Given the following trigonometry value

sin 5π/3

Since the value of π is 180 degrees, hence;

sin 5(180)/3

sin 5(60)

sin 300

This can be written as;

sin(360 - 30) = -sin30

sin(360 - 30) = -(1/2)

Hence exact value of the expression is -1/2

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According to the given data" the arithmetic mean of 6 x, 3 x , and 27 is 18" the value of x = 3.

The arithmetic mean or average, or simply the mean or average, is indeed the sum of a set of numbers multiplied by the number of numbers in the set.

One approach is to compute the arithmetic mean. To do this, add all of the values together and divide the total by the number pf values. For example, if you have a set of "n" numbers, add them together as follows: a + b + c + d, and so on. Then divide the total by "n."

If the arithmetic mean of 6 + x, 3 + x ,  27  =  18

So value of x will be as follows:

(6x + 3x + 27)

(6x + 3x + 27)/3 = 18

9x+27/3 = 18

9x +27 = 54

x = (54-27)/9

x  = 27/9

x = 3

According to the given data the value of x = 3.

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1. The private college's tuition account can earn $22,251,500 in simple interest at 2.33% in one year.

2. The account would earn $22,512,028 in one year at 2.33% interest compounded daily.

3. The difference between daily compound interest and simple interest is $260,528.

4. If the money is used to pay full scholarships, and the tuition price is $61,000 per year, 4 more students can receive full 4-year scholarships with daily compounding instead of simple interest.

Compound interest adds the earned interest to the principal and periodically compounds the new total, unlike simple interest.

1) Simple Interest:

Principal = $955 million

Period = 1 year

Interest rate = 2.33%

Simple Interest = $22,251,500 ($955,000,000 x 2.33% x 1)

2) N (# of periods) = 365 days

I/Y (Interest per year) = 2.33%

PV (Present Value) = $955,000,000

PMT (Periodic Payment) = $0

Results:

FV = $977,512,028.18

Total Daily Compound Interest = $22,512,028.18

3) The difference between daily compound interest and simple interest is $260,528 ($22,512,028 - $22,251,500).

4) Price of tuition per year = $61,000

Additional earnings from compounding interest = $260,528

Additional students to sponsor for tuition each year = 4.27 ($260,528/$61,000).

Thus, compound interest earns more for the investor than simple interest.

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Using the summation notation, the series 5+8+11+ . . . . . . +38 can be written as

[tex]\sum\limits^{12}_{n=1} {(3n+2)}[/tex]    or    [tex]24+3\sum\limits^{12}_{n=1} {n}[/tex]

Given an arithmetic series:

     5+8+11+ . . . . . . +38

The common difference (d) of each term is:

d = 8 - 5 = 3

The first term, a(1) = 5.

Hence, using the formula for the nth term of an arithmetic sequence:

a(n) = a(1) + (n-1) . d

a(n) = 5 + (n-1) . 3

a(n) = 3n + 2

Let us find n that result in a(n) = 38

a(n) = 3n + 2

38 = 3n + 2

3n = 36

n = 12

We can, then, conclude that the series  5+8+11+ . . . . . . +38 has 12 terms.

Therefore, the summation notation is:

[tex]\sum\limits^{12}_{n=1} {a(n)} = \sum\limits^{12}_{n=1} {(3n+2)}[/tex]

[tex]= \sum\limits^{12}_{n=1} {3n}+ \sum\limits^{12}_{n=1} {2}[/tex]

[tex]= 3\sum\limits^{12}_{n=1} {n}+ 2\times 12[/tex]

[tex]= 24+3\sum\limits^{12}_{n=1} {n}[/tex]

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The sets to which the numbers belong are given as follows:

32. Rational.

34. Irrational.

For this problem, we have that:

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The coordinate of point J on the number line is -9

The given parameters are

JK = KL

K = 7

L = 23

The above means that:

K = 1/2 * (J + L)

So, we have

7 = 1/2 * (J + 23)

Multiply both sides by 2

14 = J + 23

Subtract 23 from both sides of the equation

J = -9

Hence, the coordinate of point J on the number line is -9

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Answer:

It seems right to me maybe sign change?

Answer: r=6

Step-by-step explanation:

-8(4-r) =16

-32 +8r =16

8r = 16+32

8r =48

r= 48/8

r=6

The coordinate of point I is such that ΔI K L is a right-angle triangle that will be either (0,y) or (2a,y).

A triangle is a 3-sided shape that is occasionally referred to as a triangle. There are three sides and three angles in every triangle, some of which may be the same.

The sum of all three angles inside a triangle will be 180° and the area of a triangle is given as (1/2) × base × height.

Given the ΔI K L right angle triangle.

The possible coordinate of the point I will be either  (0,y) if IJ is perpendicular to JK and IK is perpendicular to JK.

Hence "The coordinate of the point I is such that ΔI K L is a right-angle triangle that will be either (0,y) or (2a,y)".

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Answer:

a) 3⁴

b)7⁵

c)10²

d)5⁷

Step-by-step explanation:

a) 3²*3², add the exponents if multiplying(subtract if division)

b) add the other 7, cause 7⁴ is 7*7*7*7 and since you're doing another *7 just add a number to the exponent

c)10⁹/10⁷ is 9-7, and take that and use it as the exponent

d)since there is no exponent but it is basically the same as b, but switched. Instead of adding an exponent you take one away

Answer:

A) 27

Step-by-step explanation:

3 x 3 = 3²

3 x 3 = 9

9 x 9

27

The Jon's age is 35 years.

We are given the product of ages of both the persons. Also, we are given Jon's age = j years and Laura's age = (j + 3) years. So, representing the given information in mathematical expression form -

j × (j + 3) = 1330

Performing multiplication on Left Hand Side of the equation to find the value of j

j² + 3j = 1330

Shifting 1330 to Left Hand Side of the equation to find the value of j

j² + 3j - 1330 = 0

Solving the mentioned quadratic equation by factorization -

j² + 38j - 35j - 1330 = 0

Further solving the quadratic equation -

j (j + 38) -35 (j + 38) = 0

(j + 38) (j - 35) = 0

j = 35, -38

The age of Jon can not be negative. So, the age of Jon will be 35 years.

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The complete question is -

Jon is 3 years younger than Laura. The product of their ages is 1,330. If j represents Jon's age and j + 3 represents Laura's age, what value of j could be Jon's age?

The real word examples of Symmetric, Transitive, and Substitution properties are {2 = 1 + 1 ⇒ 1 + 1 = 2} , {4 = 2 + 2 and 2 + 2 = 5 - 1 then 4 = 5 - 1} , { if 4 = 2 + 2 then 8×4 = 8(2+2) } respectively there is no non-example.

A decimal number is a very common number that we use frequently.

Any of the multiple sets of symbols and the guidelines for utilizing them to represent numbers are included in the Number System.

Symmetric property states that if a = b  then b = c

For example 2 = 1 + 1 ⇒ 1 + 1 = 2

Transitive property states that if a = b , b = c then a = c

For example 4 = 2 + 2 and 2 + 2 = 5 - 1 then 4 = 5 - 1

Substitution property states that if a = b then we can put b instant of an in any equation.

For example 4 = 2 + 2 then 8×4 = 8(2+2)

Hence "The example of real word on Symmetric, Transitive, and Substitution properties are {2 = 1 + 1 ⇒ 1 + 1 = 2} , {4 = 2 + 2 and 2 + 2 = 5 - 1 then 4 = 5 - 1} , { if 4 = 2 + 2 then 8×4 = 8(2+2) } respectively".

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The proving of the trigonometric function [tex]\frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A} = 2 cosec A[/tex] that LHS = RHS is shown.

So, [tex]\frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A} = 2 cosec A[/tex]:

LHS = [tex]\frac{1+\cos \mathrm{A}}{\sin \mathrm{A}}+\frac{\sin \mathrm{A}}{1+\cos \mathrm{A}}[/tex]

[tex]=\frac{(1+\cos \mathrm{A})(1+\cos \mathrm{A})+\sin \mathrm{A} \sin \mathrm{A}}{\sin \mathrm{A}(1+\cos \mathrm{A})}\\\begin{aligned}&=\frac{(1+\cos \mathrm{A})^2+\sin ^2 \mathrm{~A}}{(1+\cos \mathrm{A}) \sin \mathrm{A}} \\&=\frac{1+\cos ^2 \mathrm{~A}+2 \cos \mathrm{A}+\sin ^2 \mathrm{~A}}{(1+\cos \mathrm{A}) \sin \mathrm{A}}\end{aligned}\\[/tex]

[tex]\begin{aligned}&=\frac{1+2 \cos A+1}{(1+\cos A) \sin A} \\&=\frac{2+2 \cos A}{(1+\cos A) \sin A}\end{aligned}\\\begin{aligned}&=\frac{2(1+\cos A)}{(1+\cos A) \sin A} \\&=2 \frac{1}{\sin A}\end{aligned}\\=2cosecA[/tex]

= RHS

Hence Proved.

Therefore, the proving of the trignometric function [tex]\frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A} = 2 cosec A[/tex] is shown.

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The correct question is given below:
Prove that:  [tex]\frac{1+\cos A}{\sin A}+\frac{\sin A}{1+\cos A} = 2 cosec A[/tex]

Answer:GCF = 6

Step-by-step explanation=Notice we have these factorizations

12 = 6*2

18 = 6*3

Which shows that 6 is the greatest common factor (GCF)

So,

12+18 = 6*2+6*3 = 6*(2+3)

We use the distributive property which is a(b+c) = ab + ac

I'll leave the 2+3 part un-simplified so that you can see the two original pieces, and how the 6 can multiply with them both to get back to 12+18 again.

original sum rewrites to 6(2+3)