A spotlight shine from the ground on a wall 12 metrs away. An object 5 meteres tall moves from the sporlight to the wall at the rate of 4m/s. Whe the object is 10 meters away from the wall, the lenght of the shadow of the object on the building is (Decreasing/increasing) .... m/s

a)The initial height of the flare when it is fired is 2m.

b)The height of the flare after 1 s is 38.75m

c)The flare reaches its maximum height after 2 seconds.

d) The maximum height of the flare is 65m.

e) The flare hits the water after 8 seconds.

The given equation which is h = -5.25t² + 42t + 2, can be used to solve the following questions:

a) To get the initial height of the flare when it is fired, the value of t = 0 must be used in the given equation:

h = -5.25(0)² + 42(0) + 2h

= 0 + 0 + 2h

= 2

Therefore, the initial height of the flare when it is fired is 2m.

b) To get the height of the flare after 1 s, the value of t = 1 must be used in the given equation:

h = -5.25(1)² + 42(1) + 2h

= -5.25 + 42 + 2h

= 38.75

Therefore, the height of the flare after 1 s is 38.75m

c)The maximum height of the flare is reached when the flare is at its peak.

Therefore, the time when the flare reaches its maximum height is found by dividing -b by 2a, where the equation is in the form of y = ax² + bx + c.

The equation h = -5.25t² + 42t + 2 is in the form of y = ax² + bx + c,

where a = -5.25, b = 42, and c = 2.t = -b/2a = -42/2(-5.25)

= -2

Therefore, the flare reaches its maximum height after 2 seconds.

d) To get the maximum height of the flare, the value of t = 2 must be used in the given equation:

h = -5.25(2)² + 42(2) + 2h

= -21 + 84 + 2h

= 65

Therefore, the maximum height of the flare is 65m.

e)When the flare hits the water, the height, h, is 0.

Therefore, the time when the flare hits the water is found by setting h = 0 in the given equation and solving for t:

0 = -5.25t² + 42t + 2

Using the quadratic formula:\($$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$\)

where a = -5.25, b = 42, and c = 2.

= \(\frac{-42 \pm \sqrt{42^2 - 4(-5.25)(2)}}{2(-5.25)} $$t\)

= 8.003 or t = 1.331

Since time cannot be negative, the time when the flare hits the water is after 8 seconds. Therefore, the flare hits the water after 8 seconds.

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The ratio of the number of students who do bring their lunch to the number of students who do not is 300 : 600 or 1 : 2

Ratio shows the relationship that exists between values. It shows the proportion of one value in another value.

In other words, ratios says how much of one thing there is compared to another thing.

The middle school has 900 students.

1/3 of the students bring their lunch instead of buying lunch at school.

Therefore,

number of students that bring lunch instead of buying lunch = 1 / 3 × 900  = 300

Number of student that buy there lunch = 900 - 300 = 600

Hence,

The ratio of the number of students who do bring their lunch to the number of students who do not is 300 : 600 or 1 : 2

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Answer:

D is most possible or b either one

Step-by-step explanation:

BTW dont expect take my info to seriously but D or B

Answer:

width=9 cm

length =36cm

Step-by-step explanation:

let us suppose width of rectangle be x

then,length of rectangle =4x

given, perimeter of rectangle =90cm

we know that,

perimeter of rectangle=2(length + width)

90cm=2(4x+x)

or,90/2cm=5x

or,45cm=5x

or,(45/5)cm=x

or,x=9cm..

therefore ,width(x)=9 cm

and length (4x)=4*9=36cm

Answer:

21

Step-by-step explanation:

Since the girls have the same age, let their age be x.
Then, their average is

\(\frac{x+x}{2} = \frac{2x}{2} = x\)

Let \(S_{i}\) denote the age of 'i' girls.
Then, \(S_{8} = S_{6} + x + x - eq(1)\)

Also, we have,

\(\frac{S_{8}}{8} =15 - eq(2)\)

\(\frac{S_{6}}{6} =13 - eq(3)\)

Then eq(2):

(from eq(1) and eq(3))

\(\frac{S_{6} + 2x}{8} =15\\\\\frac{13*6 + 2x}{8} = 15\\\\78+2x = 120\\\\2x = 120-78\\\\x = 21\)

The average of the other two girls with equal age​ is 21

Answer:

6 miles

Step-by-step explanation:

because the common denominator is 12 so the man will walk 12 miles in 4 hours and the woman will walk 12 miles in 2 hours so they will meet in the middle

If point A and point B are 15 miles apart, they will meet H) 10 miles from point B.

The answer to the question is H: 10 miles.

To solve this problem, we can use the concept of relative velocity. Relative velocity is the velocity of one object as seen by another object. In this case, the man and the woman are moving towards each other, so their relative velocity is the sum of their individual velocities.

The man's velocity is 3 miles per hour and the woman's velocity is 6 miles per hour, so their relative velocity is 3 + 6 = 9 miles per hour.

We know that point A and point B are 15 miles apart, so they will meet when the distance between them is 0. The time it takes them to meet is 15 / 9 = 5/3 hours.

In 5/3 hours, the man will travel 3 * 5/3 = 5 miles and the woman will travel 6 * 5/3 = 10 miles. Therefore, they will meet 10 miles from point B.

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You cant do it on normal calculators you will need somthing like the TI-84 or CASIO FX-9860GII .

Given tan(x)=24/25 (in quadrant 1), the value of sin(2x) is 2352 / 15625.

It should be noted that tan(x) = sin(x) / cos(x)

Given tan(x) = 24/25, we can represent it as:

24/25 = sin(x) / cos(x)

cos²(x) + sin²(x) = 1

Since we're in quadrant 1, both sin(x) and cos(x) are positive. Let's solve for cos(x):

cos²(x) + (24/25)² = 1

cos²(x) + 576/625 = 1

cos²(x) = 1 - 576/625

cos²(x) = 49/625

Taking the square root of both sides:

cos(x) = sqrt(49/625)

cos(x) = 7/25

Now that we have cos(x), we can find sin(x) using the given equation:

24/25 = sin(x) / (7/25)

Multiplying both sides by (7/25):

(7/25) * (24/25) = sin(x)

168/625 = sin(x)

Now, we have sin(x) and cos(x), and we can use double angle formula to find sin(2x):

sin(2x) = 2 * sin(x) * cos(x)

Substituting the values we found:

sin(2x) = 2 * (168/625) * (7/25)

sin(2x) = (2 * 168 * 7) / (625 * 25)

sin(2x) = 2352 / 15625

Therefore, sin(2x) = 2352/15625.

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Observations consisting of pairs of variable data are required to construct a scatter plot chart. A scatter plot chart is a graphical representation of the relationship between two variables.

One variable is plotted on the horizontal axis, while the other variable is plotted on the vertical axis. Each point on the chart represents an observation or data point consisting of a pair of values for the two variables being plotted. The scatter plot chart is useful for identifying patterns, trends, and relationships between the variables. It can also be used to detect outliers or unusual observations that do not follow the general pattern of the data.

The scatter plot chart can be enhanced by adding regression lines or trend lines, which provide a visual representation of the linear relationship between the two variables. Overall, the scatter plot chart is a powerful tool for analyzing and understanding the relationship between two variables in a dataset.

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Observations consisting of pairs of variable data are required to construct a scatter chart.

A scatter chart is a graphical representation of a set of observations where one variable is plotted on the x-axis and the other variable is plotted on the y-axis. Scatter charts are useful in identifying patterns and relationships between the two variables. The strength and direction of the relationship between the variables can be determined by examining the pattern of the scatter chart. If the data points are tightly clustered around a straight line, then there is a strong linear relationship between the two variables. On the other hand, if the data points are scattered without any pattern, then there is no relationship between the two variables. Scatter charts are commonly used in scientific research, marketing analysis, and other fields where two variables need to be analyzed simultaneously.

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Step-by-step explanation:

X-2y=5 ; 2x-4y=10

• x - 2y = 5 • 2x -4y = 10

-2y = -x+5 -4y= -2x +10

y1 = ½x - 5/2 y2 = ½x -5/2

• => y1 = y2

½x -5/2 = ½x -5/2

the systems have no solution

• graph as attached

Answer:

\(\textsf{Equation 1}: \quad y=\dfrac{1}{2}x-\dfrac{5}{2}\)

\(\textsf{Equation 2}: \quad y=\dfrac{1}{2}x-\dfrac{5}{2}\)

True for all x (infinitely many solutions)

Step-by-step explanation:

Slope-intercept form of a linear equation:

\(y = mx + b\)

(where m is the slope and b is the y-intercept)

Given equations:

\(\begin{cases}x-2y=5\\2x-4y=10\end{cases}\)

To put both equations in slope-intercept form, rearrange both equations to make y the subject:

Equation 1

\(\implies x-2y=5\)

\(\implies x-2y+2y=5+2y\)

\(\implies x=5+2y\)

\(\implies x-5=5+2y-5\)

\(\implies 2y=x-5\)

\(\implies 2y \div 2=(x-5) \div 2\)

\(\implies y=\dfrac{1}{2}x-\dfrac{5}{2}\)

Equation 2

\(\implies 2x-4y=10\)

\(\implies 2x-4y+4y=10+4y\)

\(\implies 2x=10+4y\)

\(\implies 2x-10=10+4y-10\)

\(\implies 4y=2x-10\)

\(\implies 4y \div 4=(2x-10) \div 4\)

\(\implies y=\dfrac{1}{2}x-\dfrac{5}{2}\)

Therefore, both equations are the same as they have the same slope and y-intercept.

As both equations are the same, this system of equations is called an equivalent system.

To solve a system of equations by substitution, substitute one equation into the other:

\(\implies \dfrac{1}{2}x-\dfrac{5}{2}= \dfrac{1}{2}x-\dfrac{5}{2}\)

As both sides are equal, the solution is true for all x.

The solution to a graphed system of equations is the point(s) of intersection.

When graphing the given system of equations, as the lines are the same, they will be on top of each other. They are called coincident lines and have infinitely many solutions since the lines have the same slope and same y-intercept.

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Option a. z

Because in z it represents 5/8 + 3/8 + 1/8.

Look at the color in z .

Answer:

Step-by-step explanation:

d /d x [ a ^x ]

= d/ d x [ e ^ln ( a ) x ]

= e ^ln ( a ) x ⋅ d /d x [ ln ( a ) x ]

= e ^ln ( a ) x ⋅ ln ( a ) ⋅ d /d x [ x]

= e ^ln (a ) x ln ( a ) ⋅ 1

= ln ( a ) e ^ln ( a ) x

Rewrite/simplify:

= a ^x ln ( a )

Answer:

the answer is 12 degrees

The measure of angle ACB when the secant BC and the tangent AC intersect is 12°

When a secant and a tangent meet outside the circle, thwe product of the length of the secant and its external segment equals the square of the length of the tangent segment.

Therefore,

arc XA = 2(64) = 128 degrees.

Hence,

m∠C = 1 / 2 (152 - 128)

m∠C = 1 / 2 (24)

m∠ACB = 12°

Therefore, angle ACB is 12 degrees.

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Answer: (x,y)→(−x,−y)

Step-by-step explanation: The rule for a rotation by 180° about the origin is (x,y)→(−x,−y)

Answer:

69 - 52 = g

(g = 17)

Step-by-step explanation:

69 [total students] - 52 [no role] = g [those who got in]

69 - 52 = g            {simplify/combine like terms}

17 = g

g = 17

So, 17 students will be in the play

(could also be modeled as 52 + g = 69)

(this is because the only possibility for each student is to either not get in, or get in. So, if we add these two values up, we get the total number of students who auditioned).

(if you want to model the equation as such:

52 + g = 69

- 52          -52   {subtract 52 from both sides to isolate g}

g = 17)

By taking the inverse cosine of both sides, we find the solutions for t as approximately 1.8235 and 4.4590 within the range 0 ≤ t < 2π.

To solve the equation 12 sin(t) cos(t) = -3 sin(t) for t, we can first simplify the equation by dividing both sides by sin(t):

12 cos(t) = -3

Next, we can divide both sides by 12:

cos(t) = -3/12

cos(t) = -1/4

To find the values of t that satisfy this equation, we can take the inverse cosine (arccos) of both sides:

t = arccos(-1/4)

Using a calculator, we can find the values of t:

t ≈ 1.8235, 4.4590 (rounded to four decimal places)

Since the given range is 0 ≤ t < 2π, we only consider the solutions within this range. Therefore, the solutions for t are:

t ≈ 1.8235, 4.4590

In summary, the values of t that satisfy the equation 12 sin(t) cos(t) = -3 sin(t) within the range 0 ≤ t < 2π are approximately 1.8235 and 4.4590.

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Answer:

Direct proportional graphs do not need to start from the origin.

Step-by-step explanation:

Direct proportional graphs do not need to start from the origin. However, all direct proportional graphs must go through the origin.

Regardless of the value of the constant of proportionality, k, the graph of the lines all go through the point of origin, (0, 0).  

As a proof, the attached screenshot shows the graph of three equations of direct variations.  The graphed direct variation equations, y = 2x, y = -2x, and y = ¼x all go through the point of origin.

Answer:

11.20cm

11 cm=the height of the cylinder

The measure of the side length x in the right triangle is approximately 23.6 feet.

The figure in the image is a right triangle with one of its interior angle at 90 degrees.

Angle A = 29 degree

Adjacent to angle A = X

Hypotenuse = 27 ft

To solve for the missing side length x, we use the trigonometric ratio.

Note that: cosine = adjacent / hypotenuse

Hence:

cos( A ) = adjacent / hypotenuse

Plug in the given values and solve for x.

cos( 29° ) = x / 27

Cross multiplying, we get:

x =  cos( 29° ) × 27

x = 23.6 ft

Therefore, the value of x is approximately 23.6 feet.

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Answer:

it is b

Step-by-step explanation:

Answer:

-3.75

-0.55

-0.5

-0.3

-0.2

Step-by-step explanation:

-3% = -0.3

-1/2 = -0.5

-1/5 = -0.2

-3.75

-0.55

The statement implies that no matter how many peanuts and raisins are exchanged between Bob and Mark, they will always end up with an equal number of each. This suggests that the quantity of peanuts and raisins each person initially possesses is irrelevant to the final outcome.

This situation can be illustrated by the concept of a fair trade. If Bob and Mark exchange peanuts and raisins in such a way that the quantity exchanged maintains an equal number of each item for both individuals, the result will always be a balanced distribution.
For example, if Bob initially has 5 peanuts and 3 raisins, and Mark has 2 peanuts and 4 raisins, they can exchange 2 peanuts for 2 raisins. After the exchange, Bob will have 3 peanuts and 5 raisins, while Mark will have 4 peanuts and 2 raisins. The overall result is still an equal number of each item for both individuals.
This statement highlights the idea of fairness in the exchange of goods, where regardless of the initial quantities, the outcome ensures equality between the participants.

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Answer:

The polar form of (7, - 7) is (r, θ) = (7√2, - π/4)

Step-by-step explanation:

We can be 95 % confident that for smokers the pulse rate is between 3.1 and 8.9 beats per minute higher than for nonsmokers.

We are given that the total number of samples taken to observe the resting pulse rates of smokers is 26 with a mean and standard deviation of 80 bpm and 5 bpm respectively while samples taken for nonsmokers is 32 with a mean and standard deviation of 74 bpm and 6 bpm respectively.

We have independent random samples, each less than 10 % of the population, and the data appears to be approximately normal.

For smokers, n1 = 26, y1 = 80, s1 = 5

For non smokers, n2 = 32, y2 = 74, s2 = 6

t = \(((y1-y2) - H0)/ \sqrt{(s1^2/n1) + (s2^2/n2}\)

t = \(((80 - 74) - 0)/ \sqrt{(5^2/26) + (6^2/32)}\)

t = 4.15

P ≈ 0.0001

df ≈ 56

as the P-value is very small, the difference observed is not a sampling error. We have strong evidence of a difference in mean pulse rates for smokers and nonsmokers.

To find how big is that difference,

\((y1-y2)\) ± t* SE(y1-y2)

= \((80 - 74)\) ± \(t*(\sqrt{5^2/26 + 6^2/26})\)

= (3.1, 8.29)

Therefore, we can be 95 % confident that the difference in resting pulse rates between smokers and nonsmokers is between the interval of 3.1 and 8.29 beats per minute.

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Answer:

True.

Step-by-step explanation:

The circumference is the perimeter of a circle. The complete path around the circle is called the circumference.

Answer:

True

The distance around the circle makes up the circumference.

When the production manager selects a sample of items that have been produced on her production line and computes the proportion of those items that are defective, the proportion is referred to as a statistic.

The statement is true.

A statistic can be the sample mean or the sample standard deviation, which is a number computed from sample. Since a sample is random in nature therefore every statistic is a random variable (that is, it differs from sample to sample in such a way that it cannot be predicted with certainty).

Statistics are computed to estimate the corresponding population parameters.

Here the production manager selects a sample of items produced on her production line to compute the proportion of defective items, that is taken from the sample and would later be used to represent the entire bunch of items produced. Thus, the proportion can be referred to as a statistic.

Hence, the statement given is true.

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Answer:

4x+2=6x

4x-6x=-2

-2x=-2

x=2÷2

=1

In both the cases when n is even and when n is odd, the expression is odd, we can conclude that if n ∈ Z, then \(5n^2 + 3n + 7\)is odd.

To prove the proposition "if n ∈ Z, then\(5n^2 + 3n + 7\)is odd" using a direct proof with cases, we consider two cases: when n is even and when n is odd.

Case 1: n is even.

Assume n = 2k, where k ∈ Z. Substituting this into the expression, we have \(5(2k)^2 + 3(2k) + 7 = 20k^2 + 6k + 7\). Notice that \(20k^2\) and 6k are both even since they can be factored by 2. Adding an odd number (7) to an even number results in an odd number. Hence, the expression is odd when n is even.

Case 2: n is odd.

Assume n = 2k + 1, where k ∈ Z. Substituting this into the expression, we have \(5(2k + 1)^2 + 3(2k + 1) + 7 = 20k^2 + 16k + 15\). Again, notice that \(20k^2\)and 16k are even. Adding an odd number (15) to an even number results in an odd number. Therefore, the expression is odd when n is odd.

Since we have covered all possible cases and in each case, the expression is odd, we can conclude that if n ∈ Z, then 5n^2 + 3n + 7 is odd.

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The upper and lower control limits for an X-bar chart and R-chart is as follows:
UCL for X-bar chart = 171.54, LCL for X-bar chart = 120.12
UCL for R-chart = 79.87, LCL for R-chart = 0

Therefore, corrective action is needed.

The upper and lower control limits for an X-bar chart and R-chart can be determined using the following formulas:

Upper Control Limit (UCL) for X-bar chart = X-bar + A2 * R-bar
Lower Control Limit (LCL) for X-bar chart = X-bar - A2 * R-bar

Upper Control Limit (UCL) for R-chart = D4 * R-bar
Lower Control Limit (LCL) for R-chart = D3 * R-bar

Where X-bar is the average of the sample means, R-bar is the average of the sample ranges, A2 is a constant based on the sample size, and D3 and D4 are constants based on the sample size.

First, we need to calculate the average of the sample means (X-bar) and the average of the sample ranges (R-bar):

X-bar = (120 + 90 + 150 + 130 + 110 + 140 + 200 + 180 + 175 + 165 + 155 + 140) / 12 = 145.83

R-bar = ((150 - 90) + (140 - 110) + (200 - 175) + (165 - 140)) / 4 = 35

Next, we need to find the values of A2, D3, and D4 for a sample size of four:

A2 = 0.729
D3 = 0
D4 = 2.282

Now we can calculate the upper and lower control limits for the X-bar chart and R-chart:

UCL for X-bar chart = 145.83 + (0.729 * 35) = 171.54
LCL for X-bar chart = 145.83 - (0.729 * 35) = 120.12

UCL for R-chart = 2.282 * 35 = 79.87
LCL for R-chart = 0 * 35 = 0

The average of these observations is 197.25, which is above the upper control limit for the X-bar chart (171.54). Therefore, corrective action is needed.

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