PLEASE HELP 15pts
If f(4) = 7 and f'(4) = -2, f(3.97) = ?

Answer:

The constant of variation for the given quadratic equation is, 30

Step-by-step explanation:

One of the form of a quadratic equation is written as:

[tex]y = kx^2[/tex]     ....[1]

where k is the coefficient and for this case the constant of variation.

In order to obtain the answer for the given equation, we write the given equation to the form above.

[tex]0.1y=3x^2[/tex]  or

[tex]y=(\frac{3}{0.1})x^2[/tex] or

[tex]y=30x^2[/tex]

Comparing this equation with equation [1], to get the value of k;

k=30.

therefore, the  constant of variation is, 30.

Answer:

-3,2

Step-by-step explanation:

just took the test edge 2020

Answer:

a - 3b + 4

Step-by-step explanation:

The distributive property indicates that two or more terms present in a sum or subtraction multiplied by another quantity, is equal to the sum or subtraction of the multiplication of each of the terms of the sum or subtraction by the number.

So,

1/2 (2a-6b + 8) = 1/2 (2a) + 1/2 (-6b) + 1/2 (8)

Now we multiply the factor (1/2) for each of the three terms,

1/2 (2a) + 1/2 (-6b) + 1/2 (8) = (2/2)a + (-6/2)b + (8/2) = a – 3b + 4

The solution is a - 3b + 4,

So we say that 1/2(2a - 6b + 8) is equivalent to a - 3b + 4.

The distributive property is solved to the equation and the expression is represented as A = a - 3b + 4

Given data:

To apply the distributive property to the expression 1/2(2a-6b+8), distribute the 1/2 to each term inside the parentheses.

1/2(2a-6b+8) can be rewritten as:

(1/2) * 2a + (1/2) * (-6b) + (1/2) * 8

Now, let's simplify each term:

(1/2) * 2a = a

(1/2) * (-6b) = -3b

(1/2) * 8 = 4

Putting it all together, the expression becomes:

A = a - 3b + 4

Hence, on applying the distributive property to 1/2(2a-6b+8) gives the equivalent expression as a - 3b + 4.

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To find the dimensions of a rectangle with length '8 cm more than 3 times the width' and a perimeter of 64 cm, form the equation 2(3W+8) + 2W = 64. Solving for W gives the width, which can be substituted into the equation 3W+8 to give the length.

The subject of the question is algebra, specifically how to form equations based on word problems involving geometric shapes. The rectangle problem you've given can be solved using the formulas for the perimeter of a rectangle.

The perimeter 'P' of a rectangle is given by P=2(length) + 2(width). In the problem, the length 'L' is given as 8 cm more than 3 times the width 'W', or L=3W+8. The perimeter 'P' is stated as 64 cm, so substituting our equation for length into the perimeter equation, we get 2(3W+8) + 2W = 64. This equation will yield the width of the rectangle. To get the length, you simply substitute the width into the length equation.

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There are several relationships between angles in a transversal.

The measure of KNL is 102 degrees.

From the figure,

Angles KNJ and JNM are corresponding angles

So, we have:

[tex]KNJ = JNM[/tex]

This gives

[tex]3(x + 14) = 5x + 2[/tex]

Open brackets

[tex]3x + 42 = 5x + 2[/tex]

Collect like terms

[tex]3x - 5x = 2 - 42[/tex]

[tex]-2x = -40[/tex]

Divide both sides by -2

[tex]x = 20[/tex]

Angle KNL is

[tex]\angle KNL =3(x + 14)[/tex]

Substitute 20 for x

[tex]\angle KNL =3(20 + 14)[/tex]

[tex]\angle KNL =3(34)[/tex]

[tex]\angle KNL =102[/tex]

Hence, the measure of KNL is 102 degrees.

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Answer:

x = -3

y = -8

Step-by-step explanation:

I took the test

The angles of the rhombus determines the inclination of the sides of the

rhombus.

The altitude of the rhombus bisects the opposite side.

The angle of the rhombus at the vertex point of the altitude = An obtuse angle

The measure of the angles of the rhombus

The length of the sides of a rhombus are equal.

The opposite angles of a rhombus are equal.

Let x represent the length of a side of the rhombus, we have;

The bisector from the vertex of the obtuse angle = An altitude

The angle the altitude forms with the side it bisects = 90°

Therefore;

The triangle formed by the altitude, half the bisected side and the side of the rhombus adjacent to the bisected side = A right triangle

The hypotenuse side to the right triangle = The side of the rhombus

Let θ represent the angle opposite the altitude, and between the sides of

the rhombus, in the right triangle by trigonometric ratios, we have;

[tex]\displaystyle cos(\theta) = \frac{0.5 \cdot x}{x} = 0.5[/tex]

θ = arccos(0.5) = 60°

Therefore, two acute (opposite) angles of the rhombus are 60°

The two obtuse angles of the rhombus are each = (360° - 2×60°) ÷ 2 = 120°

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Answer:

–6x + 2y = 3/2 is Sandra's equation.  

Step-by-step explanation:

Given Equation of Tomas is

[tex]y=3x+\frac{3}{4}[/tex]

Solution of equation of Sandra is same as solution of Tomas Equation.

This only possible when ratio of the coefficients of the variable and constant term is equal.

So,

Multiply Tomas equation with 2.

2y= 6x + 3/2

This is one of the equation possible for Sandra's equation

Therefore, –6x + 2y = 3/2 is Sandra's equation.  

Answer:

The answer is : -15 (C)

Step-by-step explanation:

–19 = x – 4

x - 4 = -19

x = -19 + 4

x = -15

General Idea:

A cyclic quadrilateral has all its vertices on the circumference of the circle. The opposite angles of a cyclic quadrilateral are supplementary, that is it add up to get 180 degrees.

Applying the Concept:

Opposite angles B and D add up to give 180 degrees.

[tex] \angle B + \angle D=180\\ \\ 4x-20+x=180\\ Grouping \; like \; terms\\ \\ 4x+x-20=180\\ Combining \; like\; terms\\\\ 5x-20 =180\\ Adding \; 20 \; on\; both\; sides\\ \\ 5x-20+20=180+20\\ Combine\; like\; terms\\ \\ 5x=200\\ Divide\; 5\; on\; both\; sides\\ \\ \frac{5x}{5} =\frac{200}{5} \\ Simplify \; fraction\; on\; both\; sides\\ \\ x=40\\ \\ Substitute \; 40\; for\; x\; in\; the\; expression\; for\; \angle B \\ \angle B=4x-20=4(40)-20=160-20\\\\\angle B=140 [/tex]

Conclusion:

When Quadrilateral ABCD ​ is inscribed in this circle, the measure of angle B for the given diagram is [tex] 140^{\circ} [/tex]

Prior to the major issue, the woodlot had 72 more maple trees than it had after.

To solve this problem

Assuming x is a positive constant, we can represent the initial number of oak trees as 5x and the initial number of maple trees as 9x.

Prior to the problem, the ratio of maple trees to oak trees was 9:5, which produced the following equation:

[tex](9x)/(5x) = 9/5[/tex]

Later, the new ratio changed to 3:11 after new trees were planted to replace the original amount of trees. After planting, 132 oak trees were reported to have been planted. We so established the equation:

11y = 132

Solving for y:

y = 132/11 = 12

Now, we can find the new number of maple trees:

3y = 3 * 12 = 36

There were 36 maple trees when new trees were planted.

Before the problem, there were 9x as many maple trees as there were.

9x = 9 * 12 = 108

The difference between the original and new number of maple trees is:

108 - 36 = 72

So, Prior to the major issue, the woodlot had 72 more maple trees than it had after.

You would multiply 60 times 10.

Hope it helped!:)

The probability that the system functions is 1. In part b, the probability that the system functions is 2 - 2p.

To find the probability that the system functions, we need to find the probability that either c or d functions. Since c and d function independently, we can use the addition rule of probability. The probability that c functions is 1 minus the probability that c fails, which is 0.08, so the probability that c functions is 1 - 0.08 = 0.92. Similarly, the probability that d functions is 1 - 0.12 = 0.88. Since either c or d must function, we can add the probabilities of c and d functioning: 0.92 + 0.88 = 1.80 (which exceeds 1 since it's not possible for both c and d to function simultaneously). Therefore, the probability that the system functions is 1.

In part b, if both c and d have a probability p of failing, then the probability that c functions is 1 - p, and the probability that d functions is also 1 - p. Again, using the addition rule of probability, the probability that the system functions is: (1 - p) + (1 - p) = 2 - 2p.

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Answer:

a) Expected amount of winnings = -$0.084

Standard deviation = $3.00

b) Expected amount of winnings = -$0.084

Standard deviation = $1.73

c) The expected winnings for both cases is exactly the same, but the spread of the wins about the expected value/mean is low in the case (b) compared to case (a).

So, with the expected winnings using the method in (a) and (b) the same, the risk associated with betting $3 dollar per game is higher because it has a higher standard deviation. This points to higher spread of losses and wins betting $3 once than betting $1 three different times.

Step-by-step explanation:

(a) Suppose you play roulette and bet $3 on a single round. What is the expected value and standard deviation of your total winnings?

It's either one wins or not with a bet of $3.

We can then obtain the probability mass function.

Random variable X represents the amount of winnings.

- If one bets $3 and picks the right colour, the person wins $6.

X = amount of winnings = 6 - 3 = $3

probability of winning = (18/37) = 0.486

- If one bets $3 and picks the wrong colour, then the person wins $0.

X = amount of winnings = 0 - 3 = -$3

Probability of losing = (19/37) = 0.514

The probability mass function is then

X | P(X)

3 | 0.486

-3 | 0.514

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (3×0.486) + (-3×0.514) = -$0.084

Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = E(X) = -0.084

Σx²p = (3²×0.486) + [(-3)² × 0.514] = 9

Variance = 9 - (-0.084)² = 8.993

Standard deviation = √8.993 = 2.999 = $3

(b) Suppose you bet $1 in three different rounds. What is the expected value and standard deviation of your total winnings?

With a bet of $1,

It's either one wins or not with a bet of $1.

We can then obtain the probability mass function.

Random variable X represents the amount of winnings.

- If one bets $1 and picks the right colour, the person wins $2.

X = amount of winnings = 2 - 1 = $1

probability of winning = (18/37) = 0.486

- If one bets $1 and picks the wrong colour, then the person wins $0.

X = amount of winnings = 0 - 1 = -$1

Probability of losing = (19/37) = 0.514

The probability mass function is then

X | P(X)

1 | 0.486

-1 | 0.514

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (1×0.486) + (-1×0.514) = -$0.028

For 3 rounds of $1, the expected amount of winnings = 3 × -0.028 = -$0.084

Variance = Var(X) = Σx²p − μ²

μ = E(X) = -0.028

Σx²p = (1²×0.486) + [(-1)² × 0.514] = 1

Variance = 1 - (-0.028)² = 0.999216

Variance for combining 3 independent distributions = Σ λᵢ²σᵢ²

Variance of 3 rounds of $1 bets = 3[1²× var(X)] = 3 × var(X) = 3 × 0.999216 = 2.997648

Standard deviation = √(variance) = √(2.997648) = $1.73

(c) How do your answers to parts (a) and (b) compare? What does this say about the riskiness of the two games?

The expected winnings for both cases is exactly the same, but the spread of the wins about the expected value/mean is low in the case (b) compared to case (a).

So, with the expected winnings using the method in (a) and (b) the same, the risk associated with betting $3 dollar per game is higher because it has a higher standard deviation. This points to higher spread of losses and wins betting $3 once than betting $1 three different times.

Hope this Helps!!!