How is a straightedge and compass used to make basic constructions?

Answer:

∠WXY = 107°

Step-by-step explanation:

In a trapezoid, adjacent angles are supplementary, meaning they add up to 180°. ∠W and ∠WXY are adjacent, so they are supplementary. 180° - 73° = 107°.

Another way to do this is to know that there are 360° in a trapezoid. This is a right trapezoid, so two of the angles are 90°. One angle is 73°. 90° + 90° + 73° = 253°. 360° - 253° = 107°.

I hope this helps :))

The measure of angle WXY is,

= 107°

An angle is a combination of two rays (half-lines) with a common endpoint. The latter is known as the vertex of the angle and the rays as the sides, sometimes as the legs and sometimes the arms of the angle.

Given that;

Sides XY and WZ are parallel.

And, Angle W is 73 degrees and angles Y and Z are right angles.

Now, In a trapezoid, adjacent angles are supplementary, meaning they add up to 180°.

Here, ∠W and ∠WXY are adjacent, so they are supplementary.

Hence, We get;

The measure of angle WXY is,

= 180° - 73°

= 107°.

Thus, The measure of angle WXY is,

= 107°

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Answer:

$12

Step-by-step explanation:

to find sales tax you multiply the cost by the percentage

you do 150 * 8%

thats also 150 * 0.08

150 * 0.08 is 12

so you pay $12 in tax

Bret paid $12 as sales tax.

A mathematical expression is made up of terms (constants and variables) separated by mathematical operators. A mathematical equation is used to equate two expressions. Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem. In a fraction say (x/y), [y] is called denominator.

We have Bret buys a dining table that costs $150 before tax. The sales tax is 8%.

Assume that he paid $[x] as sales tax. Then, we can write -

[x] = 8% of 150

[x] = (8/100) x 150

[x] = 12

Therefore, Bret paid $12 as sales tax.

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Answer:

m>24

26>m-4

m=24

Thus, 420 fish were ultimately caught, with 6x standing in for "6 times as much."

x + y = 420; y = 6x

Thus, 420 fish were ultimately caught, with 6x standing in for "6 times as much."

The smallest and largest fish range in size from 1 cm to 18 m. 18 m then equals 18 x 100 or 1800 cm. ∴ The length of large fish exceeds that of little fish by 1800 times.

The number of parrotfish Carlos purchased is represented by the variable y, whereas the number of angelfish Carlos purchased is represented by the variable x.

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Answer:

x -2x and the y is -2y

Step-by-step explanation:

The statement that is true is that substance a is likely denser than water, whereas substance b may be less dense or have dissolved in the water.

Density is an important factor to consider when working with substances and liquids, as it can affect how they behave and interact with each other. The most likely reason for the small pile of substance a at the bottom of the first beaker is that it is denser than water. On the other hand, substance b may be less dense than water, which would cause it to float or dissolve completely. Alternatively, substance b may have dissolved in the water without leaving any visible residue.

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Answer:

x = 8.33

Explanation:

y varies directly with x if y can be calculated as a constant k times x. So:

y = k*x

If y is equal to 75 and x is equal to 25, we can calculate the value of k as:

Therefore, y = 3*x

So, to find x when y = 25, we need to replace y by 25 and solve for x as follows:

Therefore, x is equal to 8.33

The number of coins given to Amy, Ben, Carl and Debbie are 1, 3, 9 and 6.

Amy, Ben, Carl, and Debbie each have some coins. Ben has three times the number of coins that Amy has and a third of the number of coins that Carl has, and Debbie has two-thirds the number of coins that Carl has.

Let

a = Amy's coins

b = Ben's coins

c = Carl's coins

d = Debbie's coins

Therefore,

b = 3a

c = 3b

3d = 2c

The number of coins that Amy has, multiplied by the number of coins that Ben has, multiplied by the number of coins that Carl has, multiplied by the number of coins that Debbie has, is 162.

Therefore,

abcd = 162

Hence,

we need to write each variable in terms of one variable.

c = 3(3a) = 9a

3d = 2(9a) = 18a

d = 6a

Hence,

a  × 3a × 9a × 6a = 162

162a⁴ = 162

a = 1

Hence,

b = 3(1) = 3

c = 9(1) = 9

d = 6(1) = 6

Therefore, the number of coins given to Amy, Ben, Carl and Debbie are 1, 3, 9 and 6.

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Answer:

8 months tops or 7 1/2 but this in only charging monthly, so 8 months

Step-by-step explanation:

Answer:

X intercept is 3

Step-by-step explanation:

Hope this helps also I would appreciate brainliest is possible please :)

Answer:

D) \(\frac{-6\pm\sqrt{8} }{2}\)

Step-by-step explanation:

\(x^2+6x+7=0\)

\(x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\)

\(x=\frac{-6\pm\sqrt{6^2-4(1)(7)} }{2(1)}\)

\(x=\frac{-6\pm\sqrt{36-28} }{2}\)

\(x=\frac{-6\pm\sqrt{8} }{2}\) <-- correct option but not simplified fully

\(x=\frac{-6\pm2\sqrt{2}}{2}\)

\(x=-3\pm\sqrt{2}\)

Answer:

About 371

Step-by-step explanation:

This question is showing exponential growth, so use the formula a(b)^x, where a is the initial value, and b is the rate(as a decimal is working with percents). X is the time.

a=100

b=30% increase each day means 1.3 as a decimal

plug that in

100(1.3)^x

That's our exponential formula: \(100(1.3)^x\)

To find the number of algae after 5 days, plug in  x for 5.

\(100(1.3)^5\)

Evaluate (never round until the very end):  100(3.71293)

Further evaluate(now round to nearest whole number): ≈ 371

Therefore, the population is about 371 algae.

Answer:

-$407

-tax:6%

407 times 6 divided by 100

413/100

4.13

407 plus 4.13

$411.13

The approximate circumference of spaceship Earth is 160m

The formula for calculating the volume of a sphere is given as;

V = 4/3 πr³

Given that;

Now, substitute the values

65417 = 1. 3 ×3.14r³

Multiply the values

65417 = 4. 082r³

Make r the subject

r³ = 16025. 722

Find the cube root

r = 25. 2m

The circumference is expressed as;

Circumference = 2πr

Substitute the values, we have;

Circumference = 2× 3.14 × 25.2

Circumference = 160 m

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Answer:

Mean: 17.6 Median: 18 Mode: N/A Range: 27

Step-by-step explanation:

By inspecting the table, it can be seen that f(x) must be shifted left by 4 units in order to match g(x). Thus, Left by 4 units is the answer.

A quadratic function is a function of the form f(x) = ax² + bx + c, where a, b, and c are real numbers and a ≠ 0.

The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. The solutions of the equation are the points at which the graph of the function intersects the x-axis.

The functions, that are quadratic, f(x) and g(x) are described in the table. From the table, it can be seen that f(x) and g(x) are not equal to each other, as the values for each x are different. To match f(x) and g(x), one of the functions must be shifted.

By inspecting the table, it can be seen that f(x) must be shifted left by 4 units in order to match g(x).

This can be calculated by subtracting the g(x) values from the f(x) values for each x.

For example, at x = -2, the difference between f(x) and g(x) is -32.

This difference is the same for all x values, meaning that f(x) must be shifted left by 4 units to match g(x). Thus, the correct answer is Left by 4 units.

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The inflection point of f(t) is approximately t = 3.73.

(a) To determine if the function f(t) = -0.425t^3 + 4.758t^2 + 6.741t + 43.7 is increasing or decreasing, we need to find its derivative and examine its sign.

Taking the derivative of f(t), we have:

f'(t) = -1.275t^2 + 9.516t + 6.741

To determine the sign of f'(t), we need to find the critical points. Setting f'(t) = 0 and solving for t, we have:

-1.275t^2 + 9.516t + 6.741 = 0

Using the quadratic formula, we find two possible values for t:

t ≈ 0.94 and t ≈ 6.02

Next, we can test the intervals between these critical points to determine the sign of f'(t) and thus the increasing or decreasing behavior of f(t).

For t < 0.94, choose t = 0:

f'(0) = 6.741 > 0

For 0.94 < t < 6.02, choose t = 1:

f'(1) ≈ 14.982 > 0

For t > 6.02, choose t = 7:

f'(7) ≈ -5.325 < 0

From this analysis, we see that f(t) is increasing on the intervals (0, 0.94) and (6.02, ∞), and decreasing on the interval (0.94, 6.02).

(b) To find the inflection point of f(t), we need to find the points where the concavity changes. This occurs when the second derivative, f''(t), changes sign.

Taking the second derivative of f(t), we have:

f''(t) = -2.55t + 9.516

Setting f''(t) = 0 and solving for t, we find:

-2.55t + 9.516 = 0

t ≈ 3.73

Therefore, The inflection point of f(t) is approximately t = 3.73.

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The maximum value of z is 0, and the values of the decision variables are x1 = 0, x2 = 10, x3 = 0, x4 = 0.

maximize: z = c1x1 + c2x2 + ... + cnxn

subject to

a11x1 + a12x2 + ... + a1nxn ≤ b1

a21x1 + a22x2 + ... + a2nxn ≤ b2

am1x1 + am2x2 + ... + amnxn ≤ bmxi ≥ 0 for all i

In our case,

the given LP is:maximize: z = 36x1 + 30x2 - 3x3 - 4x

subject to:

x1 + x2 - x3 ≤ 5

6x1 + 5x2 - x4 ≤ 10

xi ≥ 0 for all i

We can rewrite the constraints as follows:

x1 + x2 - x3 + x5 = 5  (adding slack variable x5)

6x1 + 5x2 - x4 + x6 = 10  (adding slack variable x6)

Now, we introduce the non-negative variables x7, x8, x9, and x10 for the four decision variables:

x1 = x7

x2 = x8

x3 = x9

x4 = x10

The objective function becomes:

z = 36x7 + 30x8 - 3x9 - 4x10

Now we have the problem in standard form as:

maximize: z = 36x7 + 30x8 - 3x9 - 4x10

subject to:

x7 + x8 - x9 + x5 = 5

6x7 + 5x8 - x10 + x6 = 10

xi ≥ 0 for all i

To apply the simplex algorithm, we initialize the simplex tableau as follows:

  |  Cj   |   x5   |   x6   |   x7   |   x8   |   x9   |   x10  |    RHS  |

---------------------------------------------------------------------------

z |  0    |   0    |   0    |  36    |   30   |   -3   |   -4   |    0    |

---------------------------------------------------------------------------

x5|   0   |   1    |   0    |   1    |   1    |   -1   |   0    |    5    |

---------------------------------------------------------------------------

x6|   0   |   0    |   1    |   6    |   5    |   0    |   -1   |   10    |

---------------------------------------------------------------------------

Now, we can proceed with the simplex algorithm to find the optimal solution. I'll perform the iterations step by step:

Iteration 1:

1. Choose the most negative coefficient in the 'z' row, which is -4.

2. Choose the pivot column as 'x10' (corresponding to the most negative coefficient).

3. Calculate the ratios (RHS / pivot column coefficient) to find the pivot row. We select the row with the smallest non-negative ratio.

Ratios: 5/0 = undefined, 10/(-4) = -2.5

4. Pivot at the intersection of the pivot row and column. Divide the pivot row by the pivot element to make the pivot element 1.

5. Perform row operations to

make all other elements in the pivot column zero.

After performing these steps, we get the updated simplex tableau:

  |  Cj   |   x5   |   x6   |   x7   |   x8   |   x9   |   x10  |    RHS  |

---------------------------------------------------------------------------

z |  0    |   0    |  0.4   |  36    |   30   |   -3   |   0    |   12    |

---------------------------------------------------------------------------

x5|   0   |   1    |  -0.2  |   1    |   1    |   -1   |   0    |   5     |

---------------------------------------------------------------------------

x10|   0  |   0    |   0.2  |   1.2  |   1   |   0    |   1    |   2.5   |

---------------------------------------------------------------------------

Iteration 2:

1. Choose the most negative coefficient in the 'z' row, which is -3.

2. Choose the pivot column as 'x9' (corresponding to the most negative coefficient).

3. Calculate the ratios (RHS / pivot column coefficient) to find the pivot row. We select the row with the smallest non-negative ratio.

Ratios: 12/(-3) = -4, 5/(-0.2) = -25, 2.5/0.2 = 12.5

4. Pivot at the intersection of the pivot row and column. Divide the pivot row by the pivot element to make the pivot element 1.

5. Perform row operations to make all other elements in the pivot column zero.

After performing these steps, we get the updated simplex tableau:

  |  Cj   |   x5   |   x6   |   x7   |   x8   |   x9   |   x10  |    RHS  |

---------------------------------------------------------------------------

z |  0    |   0    |  0.8   |  34    |   30   |   0    |   4    |   0     |

---------------------------------------------------------------------------

x5|   0   |   1    |  -0.4  |   0.6  |   1    |   5   |   -2   |   10    |

---------------------------------------------------------------------------

x9|   0   |   0    |   1    |   6    |   5    |   0   |   -5   |   12.5  |

---------------------------------------------------------------------------

Iteration 3:

No negative coefficients in the 'z' row, so the optimal solution has been reached.The optimal solution is:

z = 0

x1 = x7 = 0

x2 = x8 = 10

x3 = x9 = 0

x4 = x10 = 0

x5 = 10

x6 = 0

Therefore, the maximum value of z is 0, and the values of the decision variables are x1 = 0, x2 = 10, x3 = 0, x4 = 0.

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Answer:

$234.85

Step-by-step explanation:

The gardener comes four times each month. Each month, Katrina writes a check, and $85.40 is deducted from her bank balance to pay the gardener.

4 days earning = $85.40

1 day earning = \(\dfrac{85.4}{4}=\$21.35\)

If the gardener came a total of 11 times so far, then we need to find the total amount earned by the gardener.

11 days earning = $21.35 × 11

= $234.85

Hence, he will earn $234.85.

Answer:

$234.85

Step-by-step explanation:

Answer:

Step-by-step explanation:

7 months.

Mark has 35 unique combinations to choose from, so can he co go 5 weeks with this set of clothes

Number of pants in Mark's closet = 5

Number of shirts in Mark's closet = 7

Combinations are ways to choose elements from a collection in mathematics where the order of the selection is irrelevant. Let's say we have a trio of numbers: P, Q, and R. Then, combination determines how many ways we can choose two numbers from each group.

He can choose amongst the pants in 5 ways, similarly, he can choose among the shirts in 7 ways

Total number of combinations = 5*7 = 35 unique combinations

So, if he wears one combination each day he can last 35 days or 5 weeks, without buying new clothes

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Answer:

aaaaaaaaaaaaaaaaa

Step-by-step explanation:

Answer:

151.18

Step-by-step explanation:

I can't really show my work on this, sorry.

Answer:

151.18

Step-by-step explanation:

210.58-59.40=151.18

The recurrence interval for the second largest flood on the Bow River in 1932 is approximately 1.0106 years. The probability of a flood with a discharge of 1,520 m3/s occurring next year is roughly 98.95%. When examining the 100-year trend of peak discharges, excluding major floods, there is likely a general pattern of fluctuations but with overall stability in typical peak discharge values.

Using the provided data on the highest discharge per year on the Bow River at Calgary, Canada, we can calculate the recurrence interval (R) for specific flood magnitudes and determine the probability of such floods occurring in the future. Additionally, we can examine the 100-year trend for floods on the Bow River, excluding major floods, to identify the general trend of peak discharges over time.

1) Calculating the Recurrence Interval for the Second Largest Flood (1,520 m3/s in 1932):

To calculate the recurrence interval (R) for the second largest flood, we need to determine the rank of that flood. Since there are 95 data points in total, the rank of the second largest flood would be 94 (as the largest flood, in 2013, is excluded). Applying the formula R = (n + 1) / r, we have:

R = (95 + 1) / 94 = 1.0106 years

Therefore, the recurrence interval for the second largest flood (1,520 m3/s in 1932) is approximately 1.0106 years.

2) Probability of a Flood of 1,520 m3/s Occurring Next Year:

The probability of a flood of 1,520 m3/s happening next year can be calculated by taking the reciprocal of the recurrence interval for that flood. Using the previously calculated recurrence interval of 1.0106 years, we can determine the probability:

Probability = 1 / R = 1 / 1.0106 = 0.9895 or 98.95%

Thus, the probability of a flood of 1,520 m3/s occurring next year is approximately 98.95%.

3) Examination of the 100-Year Trend for Floods on the Bow River:

To analyze the 100-year trend for floods on the Bow River while excluding major floods, we focus on the peak discharges over time. Without considering the labeled major floods, we can observe the general trend of peak discharges.

Unfortunately, without specific data on the peak discharges for each year, we cannot provide a detailed analysis of the 100-year trend. However, by excluding major floods, it is likely that the general trend of peak discharges over time would show fluctuations and variations but with a relatively stable pattern. This implies that while individual flood events may vary, there might be an underlying consistency in terms of typical peak discharges over the 100-year period.

In summary, the recurrence interval for the second largest flood on the Bow River in 1932 is approximately 1.0106 years. The probability of a flood with a discharge of 1,520 m3/s occurring next year is roughly 98.95%. When examining the 100-year trend of peak discharges, excluding major floods, there is likely a general pattern of fluctuations but with overall stability in typical peak discharge values.

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The data is represented by by quadratic function -0.28x² + 3.7x + 2.8 and it's graph attached below

Using the information given , we could create a data table which would help us make a graphical representation of the data.

Time (hours) | Snow on ground (inches)

------- | --------

0 | 3

1 | 6

2 | 9

3 | 12

6 | 13

7 | 14

8 | 15

9 | 16

10 | 12

11 | 8

Using a graphical calculator, the data yields a quadratic graph attached below.

Hence, The data is represented by the quadratic model -0.28x² + 3.7x + 2.8

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Problem 61

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===========================================================

Problem 62

T(n) = nth triangular number

T(n) = n(n+1)/2 = 0.5n(n+1)

That squares to (0.5n(n+1))^2 = 0.25n^2(n+1)^2

The next triangular number after T(n) is

T(n+1) = 0.5(n+1)(n+1+1) = 0.5(n+1)(n+2)

That squares to 0.25(n+1)^2(n+2)^2

Notice how each squared result has 0.25 and (n+1)^2 found buried in them.

Let's say A = 0.25(n+1)^2

That would mean the first result 0.25n^2(n+1)^2 becomes An^2

The second result 0.25(n+1)^2(n+2)^2 becomes A(n+2)^2

Let's add those to see what happens

An^2+A(n+2)^2

An^2+A(n^2+4n+4)

A(n^2+n^2+4n+4)

A(2n^2+4n+4)

0.25(n+1)^2*(2(n^2+2n+2))

0.5(n+1)^2(n^2+2n+1+1)

0.5(n+1)^2((n+1)^2+1)

0.5k(k+1)

We see that the result is a triangular number where k = (n+1)^2

This shows that adding the squares of consecutive triangular numbers gets us another triangular number.

A few examples

===========================================================

Problem 63

I'll borrow some of the ideas from problem 62

We found that after squaring the nth and (n+1)th triangular numbers, we got An^2 and A(n+2)^2 respectively. We let A = 0.5(n+1)^2

Subtract those expressions to get...

A(n+2)^2 - An^2

A(n^2+4n+4)-An^2

A(n^2+4n+4-n^2)

A(4n+4)

4A(n+1)

4*0.5(n+1)^2*(n+1)

(n+1)^3

This proves that the difference between the squares of consecutive triangular numbers is a perfect cube, aka a cube number.

A few examples:

===========================================================

Problem 64

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Answer:

Step-by-step explanation:

Given μA = $2.78, σA = $0.25, μB = $2.45, σB = $0.20, you want ...

The probabilities of interest are found using the CDF function of a suitable calculator or spreadsheet.

(a) P(A < $2.50) ≈ 0.1314

(b) P(B < $2.50) ≈ 59.87%

(c) P(B > $2.78) ≈ 0.0495

__

Additional comment

We note that you have provided your own answers to these questions. The answer you give for question B is not given as the percentage requested.

<95141404393>

Answer:

2?

Step-by-step explanation:

a and b are both terms

With the Hubble's constant H₀, the estimated age of the universe would be:

T = 8,332,452,617.49 years.

We know that the age of the universe is something near to the time the galaxies needed to reach their current distance:

T = D/V

And by Hubble's law, we know that:

V = H₀*D

Then we can write:

T = D/(H₀*D) = 1/H₀

So, we can say that the age of the universe is something near the inverse of Hubble's constant.

Then we have:

T = 1/(36 km/s*Mly) = (1/36)  s*Mly/km

Now we need to perform the correspondent change of units.

1 Mly = 1 million light-years

Such that:

1 ly = 9.461*10^12 km

Then 1 million light-years over km is equal to:

1 Mly/km = 1,000,000*(9.461*10^12 km)/km = 9.461*10^18

Then we can replace it:

T = (1/36) s*Mly/km = (1/36)*9.461*10^18 s

T = 2.628*10^17 s

This is the age in seconds, but we want it in years.

We know that:

1 year = 3.154*10^7 s

Then to change the units, we compute:

T = (2.628*10^17 s/3.154*10^7 s)* 1 yea

T = 8,332,452,617.49 years.

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