The perimeter of a rectangle is 96 ft. The ratio of its length to its width is 5 : 3. What are the dimensions of the rectangle?

The solution set for the inequality - 3 (x - 4) > 6(x - 1) does not include 3 as an element which is inequality is false.

Inequality is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are not equal.

We have been given inequality as:

-3(x-4) > 6(x-1)

To determine the solution set for the inequality.

Substitute x = 3 in the given inequality for each value of x with 3 and simplify both sides.

⇒ -3(x-4) > 6(x-1)

⇒ -3(3-4) > 6(3-1)

⇒ -3(-1) > 6(2)

⇒ 3 > 12

Since inequality 3 > 12 is not true,

So, for x = 3, the given inequality is false. Therefore, 3 is not included in the solution set.

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Final answer:

The standard form equation of the line through the points (-1, 1) and (1, 4) is 3x - 2y = -5.

Explanation:

The standard form equation of a line can be represented as Ax + By = C, where A, B, and C are constants. To find the equation of the line through the points (-1, 1) and (1, 4), we can use the point-slope form.

First, find the slope using the formula m = (y2 - y1)/(x2 - x1). In this case, m = (4 - 1)/(1 - (-1)) = 3/2.

Next, choose one of the given points, let's say (-1, 1), and substitute the values into the point-slope formula. y - y1 = m(x - x1). We have y - 1 = (3/2)(x - (-1)).

Simplify the equation by distributing the slope and rearranging the terms. y - 1 = (3/2)x + 3/2.

Finally, convert the equation to standard form by moving all terms to one side and multiplying by a common denominator. 3x - 2y = -5.

Answer:   4) 0.0062097

Step-by-step explanation:

Given : The length of a social media interaction is normally distributed with a mean of [tex]\mu=3[/tex] minutes and a standard deviation of [tex]\sigma=0[/tex] minutes.

Using the formula , [tex]z=\dfrac{x-\mu}{\sigma}[/tex], the z-value corresponding to x=4 will be :-

[tex]z=\dfrac{4-3}{0.4}=2.5[/tex]

Using the standard normal distribution table for z-value , the  probability that an interaction lasts longer than 4 minutes will be :-

[tex]P(z>2.5)=1-P(z\leq2.5)=1-0.9937903=0.0062097[/tex]

Hence, the probability that an interaction lasts longer than 4 minutes = 0.0062097

The statement "Five quarters are added to the bowl every week" is a correct option (A) is correct.

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

It is given that:

The function q(w)=3+5(w−1) represents the number of quarters in a bowl on week w.

q(w) = 3 + 5(w−1)

q(w) = 5w - 5 + 3

q(w) = 5w - 2

y = mx + c

m = 5

Here 5 means five quarters are added to the bowl every week.

Thus, the statement "Five quarters are added to the bowl every week" is a correct option (A) is correct.

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Answer:

width of the rectangular prism is 4.5 inches.

Step-by-step explanation:

Carltren solves for w and writes the equivalent equation as [tex]w=\frac{V}{lh}[/tex]

Now, we have to find the width of a rectangular prism that has a volume of 138.24 cubic inches a height of 9.6 inches and a length of 3.2 inches.

Thus, we have

V = 138.24 cubic inches

l = 3.2 inches

h = 9.6 inches.

Substituting these values in the above formula to find w

[tex]w=\frac{138.24}{3.2\cdot9.6}[/tex]

On simplifying, we get

[tex]w=4.5[/tex]

Thus, width of the rectangular prism is 4.5 inches.

Answer:

The distance CC' is [tex]\sqrt5units[/tex]

Step-by-step explanation:

Given the transformation T:  (x, y) (x + 2, y + 1)

we have to find the distance CC'

Let coordinate of C are (a,b).

Now, by using transformation T the coordinates of C' are (a+2,a+1)

By using distance formula,

[tex]CC'=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\= \sqrt{(a+2-a)^2+(b+1-b)^2}\\\\=\sqrt{4+1}=\sqrt5 units[/tex]

Hence, the distance CC' is [tex]\sqrt5units[/tex]

Answer:

D

Step-by-step explanation:

Answer:

3rd graph is the correct graph

Step-by-step explanation:

Given is the equation of hyperbola as

[tex]y = ± \sqrt{x^2-5}[/tex]

Square both sides and rearrange to get

[tex]y^2=x^2-5 \\x^2-y^2 =5[/tex]

Vertices are [tex](\sqrt{5} ,0) \\(-\sqrt{5} ,0)[/tex]

Asymptotes would have the same equation as hyperbola except constant term as 0

[tex]x^2-y^2 =0[/tex]

are the asymptotes

Or [tex]y = ± x[/tex] option d is right.

The weight of Stephen's dog such that Stephen has a dog that weighs 5 times as much as Ian's dog will be 60 pounds.

Determine the known quantities and designate the unknown quantity as a variable while trying to set up or construct a linear equation to fit a real-world application.

Let's assume the weight of Stephen's dog is "s" while Ion's dog is "i".

As per the given,

Stephen has a dog that weighs 5 times as much as Ian's dog.

s = 5i

Total weight, s + i = 72

Substitute, i = 72 - s into s = 5i

s = 5(72 - s)

s = 360 - 5s

s + 5s = 360

6s = 360

s = 60 pounds.

Hence "The weight of Stephen's dog such that Stephen has a dog that weighs 5 times as much as Ian's dog will be 60 pounds".

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Answer:

Its False c:

Step-by-step explanation:

Final answer:

The sum of the infinite geometric series 2 + 2/5 + 2/25 + 2/125 + ... is 2.5, determined by using the formula for the sum of a convergent geometric series, which in this case is S = 2 / (1 - 1/5).

Explanation:

The question you've asked relates to the sum of an infinite geometric series. The series given is 2 + 2/5 + 2/25 + 2/125 + ..., which is a series where each term after the first is found by multiplying the previous term by 1/5. To find the sum of this infinite geometric series, we can use the formula for the sum of a convergent geometric series, which is S = a / (1 - r), where 'S' is the sum of the series, 'a' is the first term, and 'r' is the common ratio (the factor we multiply by to get each term in the series).

In this case, the first term 'a' is 2, and the common ratio 'r' is 1/5. So our formula becomes S = 2 / (1 - 1/5), which simplifies to S = 2 / (4/5) or S = 2 * (5/4), which gives us S = 2.5. Therefore, the sum of the infinite geometric series is 2.5.

f(-4) = 20 when evaluated in the given function f(n) = n^2 - n. This result was obtained through substitution in the function followed by simplification.

In this problem, you are given a function f(n) = n^2 - n. To find the value of f(-4), you simply need to substitute -4 in place of n in the function and compute.

So,

f(-4) = (-4)^2 - (-4)

= 16 - (-4)

= 16 + 4 = 20

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Answer: [tex]c=12x+3[/tex], where c is the total cost of x pizzas.

The cost of 5 pizzas = $63

Step-by-step explanation:

Given : Cost of each pizza = $9.00

Cost of each topping =  $1.50

⇒ Cost of 2 toppings =  2 x $1.50 = $3.00

Then, the cost of each pizza having 2 toppings =  $9.00+  $3.00=  $12.00

Let x be the number of pizzas .

Then , the cost of x pizzas = 12x

Since , Delivery charge =  $3.00

Then, the total cost of ordering x pizzas( in dollars) =  Cost of x pizzas+ Delivery charge = 12x+3

Function rule to show the total cost of the pizzas if the pizza ordered has 2 toppings : [tex]c=12x+3[/tex] , where c is the total cost.

Put x= 5

[tex]c=12(5)+3=60+3=63[/tex]

Hence, the cost of 5 pizzas = $63

Answer:

SV=   41 units

QT: 21 units

Step-by-step explanation:

hope it helps:)

Given TR = 17 units, TRS = 9x - 4, VRS = 3x + 2, and QRV = 4x + 1, with x ≈ 1.583. SV ≈ 6.749 units and QT ≈ 7.332 units.

To find the lengths of SV and QT, we'll first set up equations based on the given relationships between the lengths of the segments.

Given:

- Length of TR = 17 units

- Length of TRS = 9x - 4

- Length of VRS = 3x + 2

- Length of QRV = 4x + 1

We need to find the lengths of SV and QT.

1. Length of TRS + Length of VRS = Length of TR (by the segment addition postulate)

  9x - 4 + 3x + 2 = 17

  12x - 2 = 17

  12x = 17 + 2

  12x = 19

  x = 19 / 12

  x ≈ 1.583

Now that we have found the value of x, we can find the lengths of SV and QT.

2. Length of SV = Length of VRS = 3x + 2

  Length of SV = 3(1.583) + 2

                 ≈ 4.749 + 2

                 ≈ 6.749 units

3. Length of QT = Length of QRV = 4x + 1

  Length of QT = 4(1.583) + 1

                 ≈ 6.332 + 1

                 ≈ 7.332 units

So, the lengths are:

- SV ≈ 6.749 units

- QT ≈ 7.332 units

Answer:

The answer is 30.

Step-by-step explanation:

Answer:

[tex]144.9\leq x\leq 145.5[/tex]

Step-by-step explanation:

Let x represent actual weight of object.

We have been that the weight of an object on a particular scale is 145.2 lbs. The measured weight may vary from the actual weight by at most 0.3 lbs.

[tex]|\text{Actual}-\text{Ideal}|\leq \text{Tolerance}[/tex].

Upon substituting our given values, we will get:

[tex]|x-145.2|\leq 0.3[/tex]

Applying absolute value rule [tex]|u|\leq a=-a\leq u\leq a[/tex], we will get:

[tex]-0.3\leq x-145.2\leq 0.3[/tex]

Add 145.2 on each side:

[tex]-0.3+145.2\leq x-145.2+145.2\leq 0.3+145.2[/tex]

[tex]144.9\leq x\leq 145.5[/tex]

Therefore, our required range will be [tex]144.9\leq x\leq 145.5[/tex].