Tiffany was cutting out some fabric for her friend she cut a piece that was 6 cm wide in an area of 30 cm to the power of two how long was the piece

Answer:

• Circumference of a circle I given by:

\( \dashrightarrow \: { \rm{C = \pi d}}\)

→ d is the diameter.

\({ \rm{C = \pi(3)}} \\ \\ { \boxed{ \tt{circumference = 3\pi \: cm}}}\)

Partial product multiplication involves breaking down multiplication problems into smaller, manageable parts. It is taught by demonstrating how to multiply each digit of one number by each digit of the other number and then adding the products.

Start with a simple example. For instance, you could use the problem 23 x 4. Write out the problem horizontally with the larger number on top.

Break down the numbers. Look at each digit in the larger number and multiply it by the entire smaller number. For example, 2 x 4 = 8 and 3 x 4 = 12.

Write out the partial products. Write each partial product below the original problem, aligned by the place value of the digit being multiplied. In this case, the partial products are 8 and 12.

Add up the partial products. Add up the partial products to get the final product. In this case, 8 + 12 = 20. So 23 x 4 = 92.

Practice with more examples. Once students understand the process, you can give them more complex problems to solve using partial product multiplication. Make sure to include problems that require carrying over to the next place value.

Reinforce the concept. To help reinforce the concept, you can have students create their own problems to solve using partial product multiplication.

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By the demand equation for a certain commodity, the price at which there is unitary elasticity is approximately $78.18 per unit

To find the price at which there is unitary elasticity, we need to determine the price (p) when the elasticity of demand is equal to 1.

The elasticity of demand can be calculated using the formula:

E = (dq/dp) x (p/q)

Given the demand equation: 550p + q = 86,000, we can solve for q in terms of p:

q = 86,000 - 550p

Now, we differentiate q with respect to p to find dq/dp:

dq/dp = -550

Substituting these values into the elasticity formula:

1 = (-550) x (p / (86,000 - 550p))

Simplifying the equation:

p = (86,000 - 550p) / 550

Multiplying both sides by 550 to eliminate the denominator:

550p = 86,000 - 550p

Combining like terms:

1100p = 86,000

Dividing both sides by 1100:

p = 78.18

Therefore, the price at which there is unitary elasticity is approximately $78.18 per unit (rounded to two decimal places).

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If the equation j=0. 239c, the food that contains 200 calories provides 47.8 joules of energy.

To convert the energy in calories to joules using the equation j = 0.239c, we can simply substitute the given value of c (200 calories) into the equation and solve for j:

j = 0.239c

j = 0.239 x 200

j = 47.8 joules

It's important to note that joules are the standard unit of energy in the International System of Units (SI), while calories are more commonly used in nutrition and food science.

The conversion factor between joules and calories is not exact, as it depends on the definition of the calorie used.

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Complete question is:

Joules and calories are two different units of energy. The equation j=0. 239c relates the measures, where c stands for calories and j stands for joules. If a food contains 200 calories, how many joules of energy does it provide?

well, let's move like the crab, backwards, let's start with b), then we'll do a)

b)

\({\Large \begin{array}{llll} y=ab^x \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=2\\ y=75 \end{cases}\implies 75=ab^2 \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=5\\ y=9375 \end{cases}\implies 9375=ab^5\implies 9375=ab^{2+3}\implies 9375=ab^2 b^3\)

\(\stackrel{\textit{substituting from the 1st equation}}{9375=\underset{ab^2}{(75)} b^3}\implies \cfrac{9375}{75}=b^3\implies 125=b^3 \\\\\\ \sqrt[3]{125}=b\implies \boxed{5=b}\hspace{5em}\stackrel{\textit{we know that}}{75=ab^2}\implies 75=a5^2\implies 75=25a \\\\\\ \cfrac{75}{25}=a\implies \boxed{3=a}\hspace{5em} {\Large \begin{array}{llll} y = 3(5^x) \end{array}}\)

a)

\(\begin{cases} x=0\\ y=j \end{cases}\implies j=3(5^0)\implies j=3(1)\implies j=3 \\\\\\ \begin{cases} x=4\\ y=k \end{cases}\implies k=3(5^4)\implies k=3(625)\implies k=1875\)

Answer:

x = - 15

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Sum the 3 given angles and equate to 180, that is

35 - 5 - 8x - 2x = 180 ( simplify left side )

30 - 10x = 180 ( subtract 30 from both sides )

- 10x = 150 ( divide both sides by - 10 )

x = - 15

To show that the product of the sample observations is a sufficient statistic for θ > 0 in the case of a random sample taken from a gamma distribution with parameters α = θ and β = 6, we can use the factorization theorem for sufficient statistics.

Let's denote the random sample as X₁, X₂, ..., Xₙ, where each Xi is an independent and identically distributed random variable following a gamma distribution with parameters α = θ and β = 6.

The probability density function (pdf) of a gamma distribution with parameters α and β is given by:

f(x; α, β) = (1 / (β^α * Γ(α))) * (x^(α - 1)) * exp(-x / β)

where Γ(α) is the gamma function.

The joint pdf of the random sample can be expressed as:

f(x₁, x₂, ..., xₙ; α, β) = (1 / (β^(nα) * Γ(α)^n)) * (x₁ * x₂ * ... * xₙ)^(α - 1) * exp(-(x₁ + x₂ + ... + xₙ) / β)

By the factorization theorem, the product of the sample observations, denoted as T = x₁ * x₂ * ... * xₙ, is a sufficient statistic for θ if we can express the joint pdf as the product of two functions, one depending on the sample observations T and the other on the parameter θ.

Let's rewrite the joint pdf in terms of T:

f(x₁, x₂, ..., xₙ; α, β) = (1 / (β^(nα) * Γ(α)^n)) * T^(α - 1) * exp(-(x₁ + x₂ + ... + xₙ) / β)

Now, we can separate the terms depending on T and θ:

f(x₁, x₂, ..., xₙ; α, β) = (1 / (β^(nα) * Γ(α)^n)) * T^(α - 1) * exp(-(x₁ + x₂ + ... + xₙ) / β) = g(T; α) * h(x₁, x₂, ..., xₙ; β)

Here, we can observe that g(T; α) = (1 / (β^(nα) * Γ(α)^n)) * T^(α - 1) depends only on T and α, and h(x₁, x₂, ..., xₙ; β) = exp(-(x₁ + x₂ + ... + xₙ) / β) depends only on the sample observations and β.

Therefore, we have successfully factorized the joint pdf into two functions, one depending on T and α, and the other depending on the sample observations and β. This confirms that the product of the sample observations T = x₁ * x₂ * ... * xₙ is a sufficient statistic for the parameter θ when the random sample is taken from a gamma distribution with parameters α = θ and β = 6.

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Answer:

Number 25

C. I and III.

and 26 is B

Total number of Chemistry textbooks sold is 81 and total number of Biology textbooks sold is 243

Let there are x biology textbooks sold.

No of chemistry textbooks sold = 3 the number of biology textbooks sold

ATQ,

A textbook store sold a combined total of 324 biology and chemistry textbooks in a week.

x + 3x = 324

4x = 324

x = 81

It means there are 81 biology textbooks sold and 3x = 3(81) = 243 chemistry textbooks sold.

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Answer:

The answer would be 3.8.

Step-by-step explanation:

Hope this helps!

Answer:

35\9 or 3.8

Step-by-step explanation:

\(\sf 7 \times \cfrac{5}{9} \)

Let's combine numbers in one fraction.

\(\sf \cfrac{7 \times 5}{9} \)

Multiply 7 and 5 which will give you 35.

\(\sf \cfrac{35}{9} \) or \(\sf 3.8\)

To solve the quadratic equations you provided, we can use various methods depending on the equation's form and complexity.

Here's the best method for each equation:

1. x^2 - 3x = 10:

We can rearrange the equation and set it equal to zero:

x^2 - 3x - 10 = 0

This equation can be factored as:

(x - 5)(x + 2) = 0

Therefore, the solutions are x = 5 and x = -2.

2. 4x^2 + 29x - 60 = 0:

This equation doesn't factor easily, so we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 4, b = 29, and c = -60.

Plugging these values into the formula, we get:

x = (-29 ± √(29^2 - 4 * 4 * -60)) / (2 * 4)

Simplifying further, we find:

x = (-29 ± √(841 + 960)) / 8

x = (-29 ± √1801) / 8

These are the two solutions to the equation.

3. x^2 - 6x = 0:

Here, we can factor out the common term x:

x(x - 6) = 0

This equation will be true if either x = 0 or x - 6 = 0.

Therefore, the solutions are x = 0 and x = 6.

4. 5x^2 + 5 = 9x:

Rearranging the equation, we get:

5x^2 - 9x + 5 = 0

This equation doesn't factor easily, so we can again use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 5, b = -9, and c = 5.

Plugging these values into the formula, we have:

x = (-(-9) ± √((-9)^2 - 4 * 5 * 5)) / (2 * 5)

Simplifying further, we get:

x = (9 ± √(81 - 100)) / 10

x = (9 ± √(-19)) / 10

Since the discriminant is negative, the solutions to this equation are complex numbers.

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Answer:

Option (2)

Step-by-step explanation:

HL theorem for congruence states that if hypotenuse and one leg of one right triangle are congruent to the corresponding sides of the of the right triangle, both the triangles will be congruent.

By this theorem,

ΔACB ≅ ΔFED

If AB ≅ DF [Hypotenuse]

AC ≅ EF [Leg]

By these properties,

AC = EF = 12

Or BC = DF = 28

Therefore, Option (2) will be the correct option.

Answer:

11/12

Step-by-step explanation:

change the 1/4 to 3/12 and 2/3 to 8/12 then both would have a common denominator.

Answer:

use πr2 formula

Step-by-step explanation:

Answer:

11

Step-by-step explanation:

Answer:11

Step-by-step explanation:

because you have to create the equation and solve

Answer:

4x^5 – x^4

Step-by-step explanation:

i took it and was correct

A box has candies in it: are taffy, are butterscotch, and are peppermint. The probability that Elsa will choose two taffy candies from the box is \($\frac{1}{15}$\).

There are three types of candies in a box: taffy, butterscotch, and peppermint. Each candy falls into one of these categories. Elsa would like to choose two candies for dessert.

The first candy will be chosen at random, and the second candy will be chosen at random from the remaining candies.

The probability that the first candy picked is taffy is and the possibility that the second candy picked is taffy is . The overall probability that two candies are taffy can be calculated by multiplying the two probabilities.

The probability of getting two taffy candies is:

P(Taffy, Taffy) = P(Taffy) × P(Taffy|Taffy not selected first)

                      = \(( $\frac{3}{10}$)\) \((\frac{2}{9}$)\) = \($\frac{1}{15}$\)

Therefore, the probability that Elsa will choose two taffy candies from the box is \($\frac{1}{15}$\).

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The function f(x) = ln(2 + sin(x))) is concave up for the range of x [0,2].

To discover the interval(s) on which f(x) = ln(2 + sin(x)) is concave up, compute the function's second derivative and check its sign.

To begin, compute the first derivative of f(x) with respect to x:

(1 / (2 + sin(x)) = f'(x)cos(x)

The second derivative of f(x) can therefore be found by taking the derivative of f'(x) with regard to x:

f''(x) = -(1/(2 + sin(x))(-sin(x)) cos2(x) + (1/(2 + sin(x))

When we simplify f''(x), we get:

f''(x) = -1/(2 + sin(x))²)sin(x)(sin(x)-2)

To discover the interval(s) where f(x) is concave up, we must first locate the interval(s) where f''(x) is positive. Because sin(x) might vary from -1 to 1, we must solve the inequality:

-(1/(2 + 1))^2(-1)(-1-2)≤f''(x)≤-(1 / (2 - 1))²(1) (1-2)

When we simplify this inequality, we get:

1/9 ≤ f''(x) ≤ -1/4

So, f is never negative at 0 ≤ x ≤ 2, so f is concave up in the range 0 ≤ x ≤ 2.

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Complete question - f(x) = ln(2 + sin(x)), 0 ≤ x ≤ 2 . Find the interval(s) on which f is concave up.

The number in the exponential scientific form is A = 5.729 x 10⁸

Given data ,

Let the number be represented as A

Now , the value of A is

A = 572,900,000

On simplifying , we get

From the laws of exponents , we get

The exponential form of a number is a way of representing a number using exponents, where the base is typically a number greater than 1

So , the value of A is

When you raise a number to a zero power you'll always get 1. Negative exponents are the reciprocals of the positive exponents.

A = 5.729 x 10⁸

Hence , the scientific form is A = 5.729 x 10⁸

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The given expression, (sec A x sin C - tan A x tan C)/sin B, simplifies to (x/40) x AB, where AB is the length of the side opposite the right angle in triangle ABC.

Since triangle ABC is right-angled at B, we can use the following trigonometric ratios:

sin A = opposite/hypotenuse = AC/BC

cos A = adjacent/hypotenuse = AB/BC

tan A = opposite/adjacent = AC/AB

sin C = opposite/hypotenuse = AB/BC

cos C = adjacent/hypotenuse = AC/BC

tan C = opposite/adjacent = AB/AC

Using these ratios, we can simplify the given expression as follows:

(sec A x sin C - tan A x tan C)/sin B

= [(1/cos A) x sin C - tan A x tan C]/sin B (using sec A = 1/cos A)

= [(1/cos A) x (AB/BC) - (AC/AB) x (AB/AC)]/sin B (substituting sin C and tan C)

= [(AB/BCcos A) - (AC/BC)]/sin B (simplifying)

= [(AB/BC) - (ACcos A/BCcos A)]/sin B (getting a common denominator)

= [(AB - ACcos A)/BC]/sin B (simplifying)

= [(AB - ABsin A)/BC]/sin B (substituting)

Since triangle ABC is right-angled at B, we can use the following trigonometric ratios:

sin A = opposite/hypotenuse = AC/BC

cos A = adjacent/hypotenuse = AB/BC

tan A = opposite/adjacent = AC/AB

sin C = opposite/hypotenuse = AB/BC

cos C = adjacent/hypotenuse = AC/BC

tan C = opposite/adjacent = AB/AC

Using these ratios, we can simplify the given expression as follows:

(sec A x sin C - tan A x tan C)/sin B

= [(1/cos A) x sin C - tan A x tan C]/sin B (using sec A = 1/cos A)

= [(1/cos A) x (AB/BC) - (AC/AB) x (AB/AC)]/sin B (substituting sin C and tan C)

= [(AB/BCcos A) - (AC/BC)]/sin B (simplifying)

= [(AB/BC) - (ACcos A/BCcos A)]/sin B (getting a common denominator)

= [(AB - ACcos A)/BC]/sin B (simplifying)

= [(AB - ABsin A)/BC]/sin B (substituting sin A = AC/BC)

= [AB(1 - sin A)/BC]/sin B (factoring out AB)

= [(AB/BC) x (cos A/sin A)]/sin B (using sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse)

= [cot A x (AB/BC)]/sin B (using cot A = cos A/sin A)

= (AB/BC) x (cos A/sin A) x (1/sin B) (multiplying fractions)

= (AB/BC) x (cos A/sin A) x csc B (using csc B = 1/sin B)

= (AB/BC) x cot A x csc B (using cot A = cos A/sin A)

= (BC/AB) x tan A x sin B (taking the reciprocal of both sides)

= (2BC/2AB) x (sin B/cos B) x (sin A/cos A) (multiplying and dividing by cos B and cos A)

= (BC/AB) x (2sin Bcos A)/(2sin A cos B) (simplifying)

= (BC/AB) x (sin (B + A)/sin (A + B)) (using sum and difference formulae)

= (BC/AB) x (sin C/sin 90) (since A + B + C = 90 degrees in a right-angled triangle)

= (BC/AB) x sin C

Substituting the given values, we get:

= [(2x + 20)/80] x AB

= (x/40) x AB

Therefore, the final answer is (x/40) x AB.

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Answer:

θ = 51.7

SOH-CAH-TOA

CAH: COS = ADJ/HYP   cos(θ) = 19.4/29.3

θ = 48.5

93.2 + 48.5 + x = 180

x = 38.3

90 + 38.3 + θ = 180

θ = 51.7

Step-by-step explanation:

Answer:

86????

Step-by-step explanation:

I am not certain, but wouldn't it be 86?

Radius=4mm

Height=6mm

Volume of cylinder = pir²h

3.14*4*4*6

= 301.44mm³

Answer:

60.9

Step-by-step explanation:

The difference between the rates of change for the two functions is 3.

The slope of a line indicates its direction and steepness. Finding the slope of lines in a coordinate plane can help predict whether the lines are parallel, perpendicular, or have no relationship at all without actually using a compass.

The slope of any line may be calculated using any two distinct locations on the line. The slope of a line formula is used to determine the ratio of "vertical change" to "horizontal change" between two distinct points on a line.

The slope is the rate of change of a function.

So, for function A: y = -x + 2

Compare with y = mx + b

m = -1

For function B, find slope by m = (y₂- y₁)/(x₂ - x₁)

(x₁, y₁) = (0, 0)

(x₂, y₂) = (2, -8)

So,

m = (-8 - 0)/(2 - 0)

= -8/2

= -4

The difference between the rate of change of both functions is

(- 1) - (-4)

= -1 + 4

= 3

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Answer:

please see photo attached for detailed analysis.

Carbon 14 has a half-life of 5,600 years, which means that after 5,600 years, half of the initial amount of Carbon 14 will decay. Therefore, we can use the following formula to find the amount of Carbon 14 that was present 11,200 years ago:

A = A0 * (1/2)^(t/T)

where:

A0 = initial amount of Carbon 14

A = amount of Carbon 14 remaining after time t

t = time elapsed since the initial amount was present

T = half-life of Carbon 14

We know that the fossil contains 3 g of Carbon 14, which is the amount remaining after 11,200 years. We want to find the initial amount, A0.

Let's plug in the values we have into the formula and solve for A0:

3 g = A0 * (1/2)^(11,200/5,600)

3 g = A0 * (1/2)^2

3 g = A0 * 1/4

A0 = 4 * 3 g

A0 = 12 g

Therefore, the fossil contained 12 g of Carbon 14 11,200 years ago.

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