The numerator of a fraction (which is in its simplest form) is 5 less than the denominator. If the numerator is multiplied by 2, and the denominator is increased by 16, then the fraction becomes smaller by 1/3 of it. Find the original fraction.

The equation for the nth term of the given arithmetic sequence is an = -17 + 5(n - 1), which matches answer choice C.

The given sequence is an arithmetic sequence, where each term increases by a common difference. To find the equation for the nth term (an), you start with the first term (a1 = -17) and add the common difference (d = 5) multiplied by (n - 1), since the first term is already in place.

A correct formula for an arithmetic sequence is an = a1 + (n - 1)d. Applying this to the given sequence, we have an = -17 + (n - 1)5. Simplifying this expression gives us an = -17 + 5n - 5, which further simplifies to an = 5n - 22. Thus, the correct answer is an = -17 + 5(n - 1), which corresponds to answer choice C.

The perimeter of a square measuring 5.35 cm on a side is 21.4 cm

To solve the above questions, we need to recall some of the formulas as follows:

Area of Square = (Length of Side)²

Perimeter of Square = 4 × (Length of Side)

Area of Rectangle = Length × Width

Perimeter of Rectangle = 2 × ( Length + Width )

Area of Rhombus = ½ × ( Diagonal₁ + Diagonal₂ )

Perimeter of Rhombus = 4 × ( Length of Side )

Area of Kite = ½ × ( Diagonal₁ + Diagonal₂ )

Perimeter of Kite = 2 × ( Length of Side₁ + Length of Side₂ )

Let us now tackle the problem !

Given:

Length of Side = 5.35 cm

Unknown:

Perimeter = ?

Solution:

Perimeter of Square = 4 × ( Length of Side )

Perimeter of Square = 4 × ( 5.35 )

Grade: College

Subject: Mathematics

Chapter: Two Dimensional Figures

Keywords: Perimeter, Area , Square , Rectangle , Side , Length , Width

The radius of a globe is 7.5 inches. What is the volume of the globe? Use 3.14 for pi.

Volume of the globe=[tex] \frac{4}{3} \pi r^{3} [/tex]

Volume of the globe=[tex] \frac{4}{3}*3.14*7.5^{3} [/tex]

Volume=[tex] \frac{4}{3}*3.14*421.875 [/tex]

Volume=[tex] \frac{4}{3}*1324.6875 [/tex]

Volume=[tex] \frac{5928.75}{3} [/tex]

Volume=1766.25[tex] inches^{3} [/tex]

Volume of Globe=1766.25[tex] in^{3} [/tex]

Answer: 1766.25

Step-by-step explanation: tryust

Lucy will have to pay $60 more each month.

The quantity of anything is stated as though it were a fraction of a hundred. A quarter of 100 can be used to express the ratio. Per 100 is what the term percent signifies. The symbol ‘%’ is used to symbolize it.

Lucy's employer pays 70% of the total cost of her family's health insurance plan, while Lucy pays the rest. Then Lucy will be only 30%.

If the total cost of the plan will increase from $450 per month to $650 per month.

At $450, Lucy pays 30% of $450 = 0.3 x 450 = $135

At $650, Lucy pays 30% of $650 = 0.3 x 650 = $195

Then Lucy has to pay each month more will be

⇒ $195 - $135

⇒ $60

Lucy will have to pay $60 more each month.

More about the percentage link is given below.

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The graph of [tex]y=4x+3[/tex] is a line with a slope of 4, meaning for every unit moved to the right on the x-axis, you move up four units on the y-axis. The y-intercept is 3, so the line intersects the y-axis at the point (0,3). We plot these points and draw the line through them for the complete graph.

The question asks, which of the following shows the graph of [tex]y = 4x + 3?[/tex]This is a linear equation in the format of y=mx+b, where 'm' is the slope of the line and 'b' is the y-intercept. In this case, 'm' is 4, so for every step you move to the right on the x-axis, you move four steps up on the y-axis. 'b' is 3, so the line intersects the y-axis at the point (0,3). To graph this equation, start at the point (0,3) and for every 1 unit you move to the right, move up by 4 units. Repeat this process with other values of x to find corresponding values of y. Points are then plotted on the graph and a line is drawn through them showing the relationship between x and y.

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Final answer:

The solution x=12 is correct for the problem 34+x=46, as substituting x=12 into the original equation results in 46, matching the right side. Verification by substitution is a common practice to confirm the accuracy of solutions, particularly in the case of equations with two solutions.

Explanation:

The problem provided is 34+x=46; you have also provided that x=12 is the solution. To verify if x=12 is the correct solution, we substitute x=12 into the original equation: 34+12=46. Upon calculating, 34+12 equals 46, which is indeed the right side of the equation, confirming that the solution x=12 is correct.

Moreover, in general mathematical practice, checking solutions by substituting them back into the original equation is a reliable method for verification. If a problem has two solutions, as some quadratic equations do, both solutions should be checked in this way to ensure that each solution satisfies the original equation, indicating that both are correct. For example, if we have a quadratic equation like 2x²-8=0, solving it would give us two possible values for x that can be substituted back into the equation to verify if the identity holds true for both solutions.

Final answer:

The given solution x=12 for the equation 34+x=46 is correct because when substituted back into the original equation, it results in an identity, confirming the solution's validity.

Explanation:

In mathematics, particularly in algebra, when we have an equation such as 34 + x = 46, and we are given a solution, in this case x=12, we can verify its correctness by substituting the value back into the original equation. If the equation simplifies to an identity, which is an equation that always holds true like 6 = 6, then the given solution is correct. To check if x=12 is a solution for the original problem, we perform the substitution:

34 + 12 = 46

After the substitution, the equation simplifies to:

46 = 46

This results in an identity, confirming that the given solution x=12 is indeed correct.

This process is similar to checking solutions for other types of equations, such as linear, quadratic, or higher-degree polynomials. Whether a problem has a single solution like in this case, or multiple solutions such as x=3, x=-7, each potential solution must be verified individually by substituting them back into the original equation to ensure they lead to an identity.

Answer:

Please find the attached file.

Step-by-step explanation:

Given: [tex]y=\dfrac{8}{x+3}-2[/tex]

We need to draw the graph of function with asymptote.

For horizontal asymptote:-

[tex]x\rightarrow \infty[/tex]

[tex]y=\lim_{x\rightarrow \infty}(\dfrac{8}{x+3}-2)=-2[/tex]

HA: y=-2

For vertical asymptote:-

[tex]y\rightarrow \infty[/tex]

[tex]x+3=0[/tex]

VA: x=-3

For x-intercept, Put y=0

(1,0)

For y-intercept, Put x=0

(0,2/3)

Now we will plot on graph and draw the graph.

Please see the attachment for graph.  

Number of Electoral votes for Florida is 29.

The question: North Carolina has 12 less electoral votes than Florida. To find the number of electoral votes for Florida, we can set up a subtraction equation.

Let x be the number of electoral votes for Florida.

Since North Carolina has 12 less electoral votes, the equation would be x = x + 12.

Solving for x, Florida has 29 electoral votes.

Answer:

  y = (1/2)x +1

Step-by-step explanation:

Given the two points on the line are (0, 1) and (4, 3), the line's equation can be written using the two-point form:

  y = (y2 -y1)/(x2 -x1)·(x -x1) +y1

  y = (3-1)/(4-0)·(x -0) +1 . . . . use the given point values

  y = (1/2)x +1

Answer:

y = (1/2)x +1

Step-by-step explanation:

Given the two points on the line are (0, 1) and (4, 3), the line's equation can be written using the two-point form:

 y = (y2 -y1)/(x2 -x1)·(x -x1) +y1

 y = (3-1)/(4-0)·(x -0) +1 . . . . use the given point values

 y = (1/2)x +1

Step-by-step explanation:

Answer:

A) [tex]25x^2y^{22}[/tex]

Step-by-step explanation:

The given expression is [tex](5xy^{5})^2 . (y^3)^4[/tex]

Here we have to use exponent rules and simplify the expression.

Power rule : [tex](a^m)^n = a^{mn}[/tex]

Using the above rule, we can write

[tex](5xy^{5} )^2 = 5^2.x^2.y^{5*2}[/tex]

= [tex]25.x^2.y^10[/tex]

Again using the power rule

[tex](y^{3} )^4 = y^12[/tex]

Now we have to put together this expression, we get

= [tex]25.x^2.y^{10} .y^{12}[/tex]

Now we have to use product rule.

Product rule: [tex]a^m . a^n = a^{m + n}[/tex]

Using this rule, we can simplify further

= [tex]25..x^2.y^{10 +12}[/tex]

= [tex]25x^2y^{22}[/tex]

Answer:

Option A) 25x^2y^22

Step-by-step explanation:

The function h(x, y) is given by the composition of the functions g(t) and f(x, y). With the given expressions for f and g, the function h(x, y) = g(f(x, y)) equals to (4 - xy^2 + x^2y^2) + ln(4 - xy^2 + x^2y^2).

The question asks to find the function h(x, y) from the functions g(t) and f(x, y), where g(t) = t + ln(t), and f(x, y) = 4 − xy^2+ x^2y^2. To begin with, it is necessary to substitute f(x, y) in place of 't' in function g which is g(f(x, y)), since we have h(x, y) = g(f(x, y)).

Performing the substitution, we obtain:

h(x, y) = g(f(x, y)) = (4 - xy^2 + x^2y^2) + ln(4 - xy^2 + x^2y^2).

This is your final function h in terms of x and y.

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Answer with Step-by-step explanation:

Carson buys three pencils at $0.35 each.

Cost of 1 pencil=$0.35

So, Cost of 3 pencils=$1.05    (0.35×3=1.05)

Buys a box of crayons for $2.86.

four notebooks at $1.99 each.

Cost of 1 notebook=$1.99

So, Cost of 4 notebooks=$7.96    (1.99×4=7.96)

and two pens at $1.09 each.

Cost of 1 pen=$1.09

So, Cost of 2 pens=$2.18  (1.09×2=2.18)

We have to estimate the amount he spent on school supplies.

Total Amount spent=$(1.05+2.86+7.96+2.18)

                               = $14.05

Hence, Total amount spent by Carson on school supplies is:

$14.05

Final answer:

To find the orbital period of an asteroid at 5 AU, apply Kepler's Third Law, resulting in approximately 11.18 years.

Explanation:

The question asks to find the orbital period of an asteroid with an average distance of 5 astronomical units (AU) from the Sun. We can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T, in years) is proportional to the cube of the average distance from the Sun (a, in astronomical units) for any object orbiting the Sun, expressed as-

[tex]T^2 = a^3[/tex]

Applying the formula for the asteroid in question:

Therefore, the orbital period of the asteroid is approximately 11.18 years.