Ursula has fraction 1 over 2 cup of butter. She uses fraction 1 over 4 cup of butter to make a batch of brownies.


Part A: How many batches of brownies can Ursula make with fraction 1 over 2 cup of butter? Dan has 4 over 5 cup of ice cream. How many 1 over 8 -cup servings are in 4 over 5 cup of ice cream?

Answer:

area of trapezium =1/2(a+b)*h

=1/2(7+10.4)*6.7

=1/2(17.4)*6.7

=1/2(116.58)

=116.58/2

=58.29 square cm

Answer:

58.29 cm²

Step-by-step explanation:

Hi there!

Area of a trapezoid: \(A=\displaystyle \frac{a+b}{2} *h\)

We're given that a = 7, b = 10.4 and h = 6.7. Plug these into the above:

\(A=\displaystyle \frac{7+10.4}{2} *6.7\\\\A=\displaystyle \frac{17.4}{2} *6.7\\A=8.7*6.7\\A=58.29\)

Therefore, the area of the trapezoid is 58.29 cm².

I hope this helps!

7 fishes οf the new released initially in the given expοnential pοpulatiοn grοwth. (οptiοn c)

Expοnential pοpulatiοn grοwth means that whatever the pοpulatiοn size is, it keeps increasing because the pοpulatiοn per individual remains the same, accelerating pοpulatiοn grοwth that is limited by limited resοurces. Hence, this increase in pοpulatiοn size is knοwn as expοnential pοpulatiοn grοwth.

The grοwth οf their pοpulatiοn is mοdeled by the expοnential functiοn

P(t)=\(7b^{t}\) ,where t is the time in weeks after the release and b is a pοsitive unknοwn base.

Initially that is t=0

sο fοr t=0,

p(0)=7×1 [since sοmething tο the pοwer zerο is equal tο 1]

Hence, P(0)=7

sο there were 7 fishes initially.

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Answer:

y = 1/8x + 6

Step-by-step explanation:

y = 1/8x + 6

P(X ≤ 2) = 0.636, P(X ≥ 4) = 1 - 0.990 = 0.010

Use the cumulative distribution function for the binomial distribution, 2 or fewer defective products:

P(X ≤ 2) = Σ P(X = k) for k = 0, 1, 2

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X ≤ 2) = (0.961)^22 + 22(0.039)(0.961)^21 + (22!/(2!20!))(0.039)^2(0.961)^20

P(X ≤ 2) = 0.636

To find the probability of finding 4 or more defective products in this batch, we can use the complement rule:

P(X ≥ 4) = 1 - P(X ≤ 3)

Using the cumulative distribution function as before, we can calculate:

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X ≤ 3) = (0.961)^22 + 22(0.039)(0.961)^21 + (22!/(2!20!))(0.039)^2(0.961)^20 + (22!/(3!19!))(0.039)^3(0.961)^19

P(X ≤ 3) = 0.990

Therefore, P(X ≥ 4) = 1 - 0.990 = 0.010

If the company finds 5 defective products in this batch, they should consider stopping production, as this is much higher than the expected rate of 3.9%.

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Eequation of the tangent line (b) at the given point P.

y + 5¹/² = (1/2) * (-5)⁻¹/² * (x + 5)

To find the slope of the curve at the given point P, we'll first differentiate the function y with respect to x.

Given function: y = x¹/²

Step 1: Differentiate y with respect to x.
dy/dx = (1/2) * x¹/²

Step 2: Plug in the x-coordinate of the point P into the derivative to find the slope at P.
P = (-5, -5¹/²)

dy/dx at P = (1/2) * (-5)⁻¹/²

Slope (a) = (1/2) * (-5)⁻¹/²

Now, we will find the equation of the tangent line at P.

Step 3: Use the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point P.

y - (-5¹/²) = (1/2) * (-5)⁻¹/²(x - (-5))

Step 4: Simplify the equation.

y + 5¹/² = (1/2) * (-5)⁻¹/² * (x + 5)

This is the equation of the tangent line (b) at the given point P.

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Step-by-step explanation:

h(x) = y = -16x² + 34x + 3

I assume the height is measured and calculated in feet.

anyway, to hit the ground means the height is 0.

so, we need to find the value of x (number of seconds) for which the function result is 0.

h(x) = 0

0 = -16x² + 34x + 3

a quadratic equation

ax² + bx + c = 0

has the general solutions

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = -16

b = 34

c = 3

x = (-34 ± sqrt(34² - 4×-16×3))/(2×-16) =

= (-34 ± sqrt(1156 + 192))/-32 =

= (-34 ± sqrt(1348))/-32 =

= (-34 ± 2×sqrt(337))/-32 =

= (-17 ± sqrt(337))/-16

x1 = (-17 + sqrt(337))/-16 = -0.084847484... seconds

x2 = (-17 - sqrt(337))/-16 = 2.209847484... seconds

the negative solution x1 does not make any sense, so, x2 is our valid solution.

the ball will hit the ground after about 2.2 seconds.

Answer: 2.2 seconds

Step-by-step explanation:

Since the basketball hit the ground, y should equal 0.

So -16x^2+34x+3=0

Use the quadratic formula x = (-b ± square root(b^2 - 4ac)) / 2a

In this case, a = -16, b = 34, and c = 3

So x = (-34 ± sqrt(34^2 - 4(-16)(3))) / 2(-16)

After simplifying, we get x≈-0.08 or x≈2.2

We can discard the negative value. So the answer is 2.2

Therefore, the expression for each wk in terms of the dataset (x1, y1), ..., (xM, yM) and W1, ..., Wk-1, Wk+1, ..., WM is: wk = (Σ (xij * yj) - Σ (wk' * xij^2), for j = 1 to N and k' ≠ k) / Σ xij^2, for j = 1 to N

The expression for each coefficient wk can be derived by taking the partial derivative of the OLS objective function J(w) with respect to wk and setting it equal to zero:

dJ(w)/dwk = 2 * Σ (xij * (wk * xij - yj)), for j = 1 to N

Setting this equal to zero and solving for wk, we get:

wk = (Σ (xij * yj) - Σ (wk' * xij^2), for j = 1 to N and k' ≠ k) / Σ xij^2, for j = 1 to N

Therefore, the expression for each wk in terms of the dataset (x1, y1), ..., (xM, yM) and W1, ..., Wk-1, Wk+1, ..., WM is: wk = (Σ (xij * yj) - Σ (wk' * xij^2), for j = 1 to N and k' ≠ k) / Σ xij^2, for j = 1 to N

The solution for this problem involves taking the partial derivative of the objective function J(w) with respect to each w_k and setting it to zero. This will give us a set of M normal equations, one for each coefficient w_k (1 ≤ k ≤ M).

The general expression for each w_k can be written as:

w_k = (Σ(x_i(k)y_i) - Σ(x_i(k)Σ(x_i(j)w_j)) / Σ(x_i(k)^2) for 1 ≤ j ≤ M, j ≠ k

Here, the summations run over all data points in the dataset D. The expression calculates w_k by considering the relationship between the k-th input variable x_i(k) and the output variable y_i, while taking into account the contribution of other coefficients w_j.

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Answer:

phone B

Step-by-step explanation:

Answer:

8-12a

Step-by-step explanation:

Answer:

8-(12 x 13)

Step-by-step explanation:

12 x 13= 156

156-8=142

Answer:

Step-by-step explanation:

Explanation: A linear pair of angles is formed when two lines intersect. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. The measure of a straight angle is 180 degrees, so a linear pair of angles must add up to 180 degrees.

Answer:

A

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

Sum means to add so answer A is correct

Answer:

5/10=50%

Step-by-step explanation:

Answer:

50%

Step-by-step explanation:

Answer:

48 cups of sugar.

Step-by-step explanation:

12 cups of sugar for 8 cups of lemonade then I let 32 cups of lemonade corresponding with x cups of sugar, then 12/8 = x/ 32, then 8x = 32 * 12, then solve for x get 8x = 384, x =48.

Answer:

32÷8= 4 cups of sugar...........

To create a validation rule in Salesforce that automatically selects grade A when the percentage is set to 90, you can use the ISPICKVAL function. This function allows you to check the selected value of a picklist field and perform actions based on the value. By using ISPICKVAL in the validation rule, you can ensure that the grade field is populated with A when the percentage field is set to 90.

To implement this validation rule, follow these steps:

Go to the Object Manager in Salesforce and open the Student object.

Navigate to the Validation Rules section and click on "New Rule" to create a new validation rule.

Provide a suitable Rule Name and optionally, a Description for the rule.

In the Error Condition Formula field, enter the following formula:

AND(ISPICKVAL(Percentage__c, "90"), NOT(ISPICKVAL(Grade__c, "A")))

This formula checks if the percentage field is selected as 90 and the grade field is not set to A.

In the Error Message field, specify an appropriate error message to be displayed when the validation rule fails. For example, "When percentage is 90, grade must be A."

Save the validation rule.

With this validation rule in place, whenever a user selects 90 in the percentage field, the grade field will automatically be populated with A. If the grade is not set to A when the percentage is 90, the validation rule will be triggered and display the specified error message.

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Answer:

12 photos fit on page = 1

432 photos fit on page = 432/12

= 36 pages

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

Answer:

36 pages

Step-by-step explanation:

There are 12 photos per page, and Robert has 432 photos. So, we can divide the number of photos he has by the number of photos that fits on each page. By doing so, you'll calculate the number of pages he needs. This calculation is shown below:

\(\frac{432 \ photos}{1} *\frac{1 \ page}{12 \ photos} = 36 \ pages\)

You can see that the units for photos cancel out, resulting in units for pages.

So, Robert will need 36 pages in order to fit all 432 photos into his photo album.

The integral ∫∫R5(x+y)ex2−y2dA, where R is the rectangle enclosed by the lines x−y=0,x−y=3,x+y=0, and x+y=4, can be evaluated by making the change of variables u = x + y and v = x - y, which gives us the Jacobian of the transformation as |J| = 1/2.

To evaluate the integral using the change of variables, we first need to find the bounds of integration for the new variables u and v. Using the equations of the lines that bound the rectangle R, we can rewrite them in terms of u and v as v = ±(3 - u) and v = ±u. These equations represent the four lines that form the new rectangle R' in the uv-plane.

The integral can now be rewritten as ∫∫R'5(u/2)eu2/2dvdu. The limits of integration for v are from -(3 - u) to (3 - u) for the bottom and top sides of the rectangle R', and from -u to u for the left and right sides. The limits of integration for u are from 0 to 4.

After integrating with respect to v, we get ∫(3-u)^u(-3+u)5(u/2)eu2/2dvdu + ∫u^-u^3 5(u/2)eu2/2dvdu. These integrals can be solved by using the substitution method. Finally, we get the answer as 17(e^16 - e^(4/3))/12.

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The solution to the systems of equations graphically is (2, 4)

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

y = 2x

y = -x + 6

Next, we plot the graph of the system of the equations

See attachment for the graph

From the graph, we have solution to the system to be the point of intersection of the lines

This points are located at (2, 4)

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Using the ANOVA model, you can determine if the surface type has a significant impact on the speed of guineas.

Given that you are comparing the speed of guineas across three different surface types (grass, turf, and concrete), you would use an Analysis of Variance (ANOVA) model to analyze this data.

An ANOVA model allows you to compare the means of the speeds for each group (grass, turf, and concrete) and determine if there are any significant differences between them. The model takes into account the variability within each group and the variability between the groups to determine if the differences observed are due to chance or if they are statistically significant.

Here are the steps to perform an ANOVA analysis:

1. Collect the speed data for each guinea in the three groups (grass, turf, and concrete).
2. Calculate the means of the speeds for each group.
3. Calculate the overall mean of the speeds for all groups combined.
4. Calculate the Sum of Squares Within (SSW), which measures the variability within each group.
5. Calculate the Sum of Squares Between (SSB), which measures the variability between the groups.
6. Calculate the Mean Squares Within (MSW) and Mean Squares Between (MSB) by dividing the respective sums of squares by their degrees of freedom.
7. Calculate the F-statistic by dividing MSB by MSW.
8. Compare the F-statistic to the critical value from the F-distribution table based on the chosen level of significance (e.g., 0.05) and the degrees of freedom for the numerator and denominator.
9. If the F-statistic is greater than the critical value, you can conclude that there are significant differences between the groups' mean speeds, and further analysis can be conducted to determine which specific groups differ.

Using the ANOVA model, you can determine if the surface type has a significant impact on the speed of guineas.

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Answer:

A.7

Reasoning:

\(4<7-2<8\)

\(4<5<8\)

The other numbers are too little, thus making the inequality untrue.

Answer:

To determine if the point (2, -8) is a solution to the system of inequalities:

y < x - 1 ----(1)

y > -x + 2 ----(2)

We need to check if the point (2, -8) satisfies both inequalities.

First, let's check inequality (1):

y < x - 1

Substituting x = 2 and y = -8, we get:

-8 < 2 - 1

-8 < 1

The inequality is true.

Now, let's check inequality (2):

y > -x + 2

Substituting x = 2 and y = -8, we get:

-8 > -2 + 2

-8 > 0

The inequality is false.

Since the point (2, -8) satisfies inequality (1) but not inequality (2), it is not a solution to the system of inequalities.

Therefore, the point (2, -8) is not a solution to the system of inequalities y < x - 1 and y > -x + 2.

Answer:

Triangle ABC is congruent to triangle DFE.

Answer:

A I believe good luck

Step-by-step explanation:

To find the real and imaginary parts of a complex number, write it in the form a + bi, where a is the real part and b is the imaginary part.

To find the real and imaginary parts of a complex number, you can use the following steps:1. Write the complex number in the form a + bi, where a is the real part and b is the imaginary part.

2. Identify the coefficient of the imaginary unit, "i." This coefficient is the value of "b" in the complex number.

3. The real part of the complex number is given by "a," and the imaginary part is given by "b."

For example, let's consider the complex number z = 3 + 2i.The real part, denoted as Re(z), is 3, and the imaginary part, denoted as Im(z), is 2.Therefore, Re(z) = 3 and Im(z) = 2.By following these steps, you can easily determine the real and imaginary parts of any complex number.

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b is a linear combination of the given vectors a1, a2, and a3.

To determine whether or not b is a linear combination of the given vectors a1, a2, and a3, we can set up the following system of equations: a1x + a2y + a3z = where x, y, and z are scalar constants.

We can then use matrix operations to solve for x, y, and z. If we are able to find a solution where all three constants are real numbers, then b is a linear combination of a1, a2, and a3. If there is no solution or if at least one constant is a complex number, then b is not a linear combination of a1, a2, and a3.

Using the given vectors and b, we can set up the following augmented matrix:1 0 2 -5-2 5 0 11 5 8 -7

To solve for x, y, and z, we can use row operations to put the matrix in row echelon form. Here are the steps: R2 + 2R1 -> R2 (to eliminate the -2 in row 2, column 1)R3 - R1 -> R3 (to eliminate the 1 in row 3, column 1)1 0 2 -5 0 5 4 1 0 5 10 -12

Now we have a system of three equations in three variables:x + 2z = -5(5)y + 4z = 1(10)x + 4y - 3z = -12We can use back-substitution to solve for z, then y, then x.

We'll start with the third equation:x + 4y - 3z = -12x + 4y - 3(1) = -12x + 4y = -9Now we'll use the second equation to solve for y:5y + 4z = 1y = (-4/5)z + (1/5)

Substitute this into the first equation:x + 2z = -5x + 2(-4/5)z + (2/5) = -5x = (22/5) - (6/5)zSubstitute this and y into the original equation:a1x + a2y + a3z = b(1)(22/5 - 6/5z) + (0)(-4/5z + 1/5) + (2)(z) = -5(1) + (-2)(1) + (1)(-7)22/5 - 6/5z + 2z = -5 - 2 - 7

Multiplying everything by 5 gives us:22 - 6z + 10z = -70z = -24/4 = -6Now that we have found z, we can substitute it back into our equation for y:y = (-4/5)(-6) + (1/5) = 5

And now we can substitute z and y into our equation for x:x = (22/5) - (6/5)(-6) = 10Thus, we have found a solution for x, y, and z: x = 10, y = 5, z = -6.

Therefore, we can conclude that b is a linear combination of the given vectors a1, a2, and a3.

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Answer: 5 dollars

Step-by-step explanation:

75/15 = 5

∫ (4x^5 - 6csc^2(x) + 10e^x - 200) dx = (4/6)x^6 + 6cot(x) - 10e^x - 200x + C, where C is the constant of integration using indefinite integral.

We need to use the following integration rules:

∫csc2(x) dx = -cot(x) + C, where C is the constant of integration.

∫e^x dx = e^x + C, where C is the constant of integration.

Using these rules, we have:

∫(4x^5 - 6csc^2(x) + 10e^x - 200) dx

= (4/6)x^6 + 6cot(x) + 10e^x - 200x + C, where C is the constant of integration.

Therefore, the indefinite integral is:

(4/6)x^6 + 6cot(x) + 10e^x - 200x + C

where C is the constant of integration.

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The graph will cross the x-axis if the multiplicity of the real root is odd.

What is polynomial?

In arithmetic, a polynomial is an expression consisting of indeterminates and coefficients, that involves solely the operations of addition, subtraction, multiplication, and positive-integer powers of variables.

Main body:

For polynomials, the graph will cross the x-axis if the multiplicity of the real root is odd, and just touch the x-axis if the multiplicity of the real root is even. (The multiplicity of the root is the number of times it occurs as a root)

(a) y=(x+1)^2(x-2) The graph crosses at x=2 (multiplicity 1) but touches at x=-1 (mulitplicity 2)

(b) y=(x-4)^3(x-1)^2 The graph crosses at x=4 (multiplicity 3) but touches at x=1 (m=2)

(c) y=(x-3)^2(x+4)^4 The graph touches at x=3 and x=-4 as the multiplicities are both even.

The graphs: (a) black, (b) red, (c) green

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Answer:

(f+g)(2)=5

Step-by-step explanation:

Subsitute 2 in for x in both functions. Then add the answers of the functions together.

An ordered pair is a solution of an equation in two variables if replacing the variables by the coordinates of the ordered pair results in a true statement.

In mathematics, an equation in two variables represents a relationship between two quantities. An ordered pair consists of two values, typically denoted as (x, y), that represent the coordinates of a point in a two-dimensional plane.

When these values are substituted into the equation, if the equation holds true, then the ordered pair is considered a solution or a solution set to the equation. This means that the relationship described by the equation is satisfied by the values of the ordered pair. In other words, the equation is true when evaluated with the values of the ordered pair.

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