assume that the diameter d of a circle is a differentiable function of t, and let c be the circumference of the circle. find an equation that relates d times c divided by d times t and times d divided by d times t. c = πd d times c divided by d times t is equal to pi times d d times c divided by d times t is equal to pi times the quantity d times d divided by d times t end quantity d times c divided by d times t is equal to pi times d times the quantity d times d divided by d times t end quantity

The fraction is equal to 3/7

the sum of numerator and denominator is equal to 100

Let the numerator = x

Since, Numerator + Denominator = 100

x + Denominator = 100

Denominator = 100 - x

Thus, the fraction is in the form of;

Since it is given that the fraction is equal to 3/7

So,

Simplify the expression for x;

So, the numerator is 30

Since, Denominator = 100 - x

Substitute x = 30

Denominator = 100 - x

Denominator = 100 - 30

Denominator = 70

Thus, we have numerato = 30 denominator = 70

So, the resulting fraction is given as;

Answer: 30/70

Answer:

16y^5

Step-by-step explanation:

hope this helps with the work

Answer:

the answer is 16\(y^{5}\)

Step-by-step explanation:

Answer:

\( \boxed{\sf Length \ of \ the \ rectangle = 16 \ in} \)

Given:

Perimeter of rectangle = 48 in

Length = Twice the width

To Find:

Length of the rectangle

Step-by-step explanation:

Let width of the rectangle be 'w'.

So,

Length of the rectangle = 2w

\(\sf \implies Perimeter \ of \ rectangle = 2(Length + Width \\ \\ \sf \implies 48 = 2(2w + w) \\ \\ \sf \implies 48 = 2(3w) \\ \\ \sf \implies 48 = 6w \\ \\ \sf \implies 6w = 48 \\ \\ \sf \implies \frac{ \cancel{6}w}{ \cancel{6}} = \frac{48}{6} \\ \\ \sf \implies w = \frac{48}{6} \\ \\ \sf \implies w = \frac{8 \times \cancel{6}}{ \cancel{6}} \\ \\ \sf \implies w = 8 \: in\)

Width of the rectangle (w) = 8 in

Length of the rectangle = 2w

= 2 × 8

= 16 in

Answer:

11.9a − 3

Step-by-step explanation:

Answer:

Step-by-step explanation:

24 : 30 = 4 : 5

44 : 99 = 4 : 9

108 : 36 = 3 : 1

33 : 9 = 11 : 3

49 : 14 = 7 : 2

The number of ceiling boards in the room is 5000

From the question, we have the following parameters that can be used in our computation:

Measurement = 5 cm by 10 cm

Room = square of 5 m dimension

The area of the room is calculated as

Area = (5 m)²

When evaluated, we have

Area = 25 m²

The area of the board is

Area = 5 cm * 10 cm

So, we have

Area = 50 cm²

So, we have

Boards = (25 m²)/(50 cm²)

Evaluate the quotient

Boards = 5000

Hence, the number of boards is 5000

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Answer:

25 students in all

Step-by-step explanation:

15 x 5/3

=25

Based on the text, the primary source of primate endangerment today is habitat destruction.

Habitat destruction is considered the primary source of primate endangerment based on various factors mentioned in the text. Primates, like many other species, heavily rely on their natural habitats for survival, including food, shelter, and breeding. However, human activities such as deforestation, urbanization, and land conversion for agriculture or infrastructure development have significantly reduced and fragmented primate habitats.

Habitat destruction leads to the loss of critical resources and disrupts the ecological balance necessary for primate populations to thrive. As their habitats shrink, primates face increased competition for limited resources, reduced access to food and water sources, and decreased opportunities for successful reproduction and rearing of offspring. This loss of habitat also exposes primates to increased vulnerability to predation, disease transmission, and human-wildlife conflicts.

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Answer:

Step-by-step explanation:

Since the length, i, of the rectangle is greater than its width, we can write:

i > w

The formula for the perimeter of a rectangle is:

P = 2(i + w)

We know the perimeter is at least 30 inches, so we can write:

2(i + w) ≥ 30

Simplifying the inequality, we get:

i + w ≥ 15

Now we can substitute i > w into the inequality:

w + w ≥ 15

2w ≥ 15

w ≥ 7.5

Therefore, the range of possible widths for the rectangle is:

w ≥ 7.5

The length of BC in this problem is given as follows:

BC = 3.6.

Two triangles are defined as similar triangles when they share these two features listed as follows:

The similar triangles for this problem are given as follows:

ABC and ADE.

Hence the proportional relationship for the side lengths is given as follows:

3/5 = BC/6.

Applying cross multiplication, the length BC is given as follows:

BC = 6 x 3/5

BC = 3.6.

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Answer:

I think it’s B (I’m not sure if I’m right)

Step-by-step explanation:

Answer: $140

Step-by-step explanation:

Answer:

140

Step-by-step explanation:

i did it before

Answer: 80 x  −  60.55

Answer:

a) Low: 45 Q1=80 Q2=89 Q3=94 High=100 b) Look at attached image

Step-by-step explanation:

Organized numbers: 45,70,72|,80,|82,85,88|,89,|90,92,92|,94,|98,100,100

mark medians

Graph created in Desmos

Answer:

9x(3x-2) (3x+2)

Step-by-step explanation:

\(81x^{3}-36x\)

9(9x³ - 4x)

9x(9x² - 4)

Since both terms are perfect squares factor using the difference of squares formula,  a² - b² = (a + b) (a - b)

Therefore, 9x(3x-2) (3x+2)

Answer:

81z^3-36z = 9z(3z-2)(3z+2)

Step-by-step explanation:

(81z^3-36z)=9z(9z^2-4)

9z(9z^2-4)=9z(3z-2)(3z+2)

81z^3-36z=9z(3z-2)(3z+2)

81z^3-36z=9z(3z-2)(3z+2)

then

Answer:

X is a variable and can represent any number that is unknown

The statement "There is a strong positive correlation between studying and grades" best describes the meaning of a correlation coefficient of 0.93. The correct answer is A.

The correlation coefficient measures the strength and direction of the linear relationship between two variables. In this case, a correlation coefficient of 0.93 indicates a strong positive correlation between studying and grades.

A positive correlation means that as the value of one variable (study time) increases, the value of the other variable (grades) also tends to increase. A correlation coefficient close to +1 indicates a strong positive relationship, suggesting that an increase in study time is associated with higher grades. e correct answer is A.

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The  percentage of each types of s. aureus

Total number of individuals = 1000

Number of individuals susceptible (fs) to methicillin = 900

The percentage = (Number of individuals susceptible (fs) to methicillin / Total number of individuals) × 100

= (900/1000) × 100

= 90%

90 are somewhat susceptible (ss) to methicillin

The percentage = (90 /1000) × 100

= 9%

10 are killed only by a full treatment with methicillin (ft)

The percentage = (10/1000) × 100

= 1%

Therefore, the percentage of each types of s. aureus are 90%, 9% and 1%

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Answer:

I can only answer ii.

A=Lxwxh

=6mx3mx4m

=18mx4m

A=40msquared .

A=Lxwxh

40msquared = 6mx3mxh

40msquared =18msquaredxh

40/18msquared=18xh/18msquared

131/3=h.

that's what I thought

Step-by-step explanation:

I think I should first find the area

of the figure.then I will use the area to find the height.

Answer:

A=75, B=60, C=45

Step-by-step explanation:

5x+4x+3x=180

12x=180

x=15

plug them in and get

A=5(15)=75

B=4(15)=60

C=3(15)=45

Answer:

8x

Step-by-step explanation:

Because you are adding the (x+3) there you can immediately you the association property to simplify

(2x + 5x + x) + (3 - 3)

8x + 0 =

8x

The correlation coefficient between history and math exam scores is of -0.1321, which means that there is a weak negative correlation between history grades and math grades.

Using a calculator, inserting the history and math scores, the coefficient is of -0.1321, which means that there is a weak negative correlation between history grades and math grades.

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7.2 is the unknown value of x from the triangle.

The given diagram is a triangle. We need to determine the measure of the value 'x'.

Using the trigonometry identity'

tan theta = opposite/adjacent

tan 50 = x/6

Cross multiply to have:

x = 6tan50

x = 6(1.1918)

x = 7.1505

Hence the value of x to the nearest tenth is 7.2

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Answer:

\( = \frac{57}{2} \\ = 28.5\)

Answer:

Step-by-step explanation:

The volume is \(V = \frac{1}{3} b^{2} h\)

Known facts:

Let us plug in the known value into the pyramid

   \(v=\frac{1}{3} * 5^{2} *6 = 50\) cubic centimeters

The volume of a square pyramid is 50 cubic centimeters!

Hope that helps!

The area of the polygon is solved to be  1044 squared cm

The area of the composite polygon is solved by dividing the object into two sections. Then adding up the areas

Section 1 has dimensions:

length * width = 46 * 14 = 644

section 2 has dimensions:

length = 46 - 21 = 25

width = 30 - 14 = 16

Area = 25 * 16 = 400

Area of the composite figure

section 1  + section 2

= 644 + 400

= 1044 squared cm

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The given equation, y = -1.7x + 8.5, is not in standard form with integer coefficients. The standard form of a linear equation is Ax + By = C, where A, B, and C are integers and A is positive.

To convert the given equation to standard form, we need to eliminate the decimal coefficient. We can do this by multiplying both sides of the equation by 10 to clear the decimal: 10y = -17x + 85 Next, we want the coefficient of x to be positive, so we can multiply both sides of the equation by -1: -10y = 17x - 85

Now, we can rearrange the terms to match the standard form: 17x - 10y = 85 So, the equation of the given line in standard form with integer coefficients is 17x - 10y = 85. To convert the given equation to standard form, we need to eliminate the decimal coefficient.

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It looks like the boundaries of \(R\) are the lines \(y=\dfrac23x \) and \(y=3x\), as well as the hyperbolas \(xy=\frac23\) and \(xy=3\). Naturally, the domain of integration is the set

\(R = \left\{(x,y) ~:~ \dfrac{2x}3 \le y \le 3x \text{ and } \dfrac23 \le xy \le 3 \right\}\)

By substituting \(x=\frac uv\) and \(y=v\), so \(xy=u\), we have

\(\dfrac23 \le xy \le 3 \implies \dfrac23 \le u \le 3\)

and

\(\dfrac{2x}3 \le y \le 3x \implies \dfrac{2u}{3v} \le v \le \dfrac{3u}v \implies \dfrac{2u}3 \le v^2 \le 3u \implies \sqrt{\dfrac{2u}3} \le v \le \sqrt{3u}\)

so that

\(R = \left\{(u,v) ~:~ \dfrac23 \le u \le 3 \text{ and } \sqrt{\dfrac{2u}3 \le v \le \sqrt{3u}\right\}\)

Compute the Jacobian for this transformation and its determinant.

\(J = \begin{bmatrix}x_u & x_v \\ y_u & y_v\end{bmatrix} = \begin{bmatrix}\dfrac1v & -\dfrac u{v^2} \\\\ 0 & 1 \end{bmatrix} \implies \det(J) = \dfrac1v\)

Then the area element under this change of variables is

\(dA = dx\,dy = \dfrac{du\,dv}v\)

and the integral transforms to

\(\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \int_{\sqrt{2u/3}}^{\sqrt{3u}} \frac{dv\,du}v\)

Now compute it.

\(\displaystyle \iint_R 9xy \, dA = \int_{2/3}^3 \ln|v|\bigg|_{v=\sqrt{2u/3}}^{v=\sqrt{3u}} \,du \\\\ ~~~~~~~~ = \int_{2/3}^3 \ln\left(\sqrt{3u}\right) - \ln\left(\sqrt{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln(3u) - \ln\left(\frac{2u}3\right) \, du \\\\ ~~~~~~~~ = \frac12 \int_{2/3}^3 \ln\left(\frac{3u}{\frac{2u}3}\right) \, du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \int_{2/3}^3 du \\\\ ~~~~~~~~ = \frac12 \ln\left(\frac92\right) \left(3-\frac23\right) = \boxed{\frac76 \ln\left(\frac92\right)}\)

The spring will stretch 1.5 times further when the 0.60 kg block is added to the 1.2 kg block.

To determine how much further the spring will stretch when a 0.60 kg block is added to the 1.2 kg block, we need to consider Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

Let's assume that the spring is initially stretched by a certain amount x when only the 1.2 kg block is attached. The force exerted by the spring is given by F = kx, where k is the spring constant.

When the 0.60 kg block is added, the total mass becomes 1.2 kg + 0.60 kg = 1.8 kg. The force exerted by the spring remains the same, but the total mass affects the acceleration of the system according to Newton's second law, F = ma.

Using F = ma, we can find the acceleration of the system: F = (1.8 kg)(9.8 m/s^2) = 17.64 N.

Since F = kx, we can rearrange the equation to find x: x = F/k.

Now, if we assume the spring constant k remains constant, the displacement x will increase by a factor of 1.8 kg/1.2 kg = 1.5.

Therefore, the spring will stretch 1.5 times further when the 0.60 kg block is added to the 1.2 kg block.

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