MATHEMATICS HIGH SCHOOL

A deep sea diver descends below the surface of the water at a rate of 60 feet each minute. What is the depth of the diver after 10 minutes.

Answers

Answer 1

Answer:

Answer:

600 feet

Step-by-step explanation:

60 feet per minute

m = minute

60m

60(10)

600 feet

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What is the length of a room with a width of 10 feet
Which is the solution set of the inequality 15y-9 < 36y>9/5y<9/5y < 3y>3
X^2-5x-24=0 3x^2+x-4=0 Solve by factoring or using the quadratic formula
I need to find how many solutions does the system have.

A polynomial function has a root of -5 with multiplicity 3, a root of 1 with multiplicity of 2, and a root of 3 with multiplicity 7. If the function has a negative leading coefficient and is of even degree which statement about the graph is true?

a) function is positive on (-∞,-5)
b) fuction is negative on (-5,3)
c) function is positive on (-∞,1)
d) function is negative on (3,∞)

Answers

Answer:

d) function is negative on (3,∞)

Step-by-step explanation:

The even degree and negative leading coefficient tell you that the function is negative as x ⇒ ±∞. (Selections A and C cannot be correct.)

The odd multiplicity tells you the function crosses the x-axis at x=-5 and x=3, so will be non-negative between those values. (Selection B cannot be correct.)

The function is negative on (3, ∞).

Answer:

The graph of the function is positive on (-co, -5).

The graph of the function is negative on (3,co).

Step-by-step explanation:

We know that the roots are in: -5, 1 and 3.

and after 3, the graph is in the negative side, so between 1 and 3 the graph must be in the positive side, between -5 and 1 the graph must be in the negative side, and between -inifinity and -5 the graph must be in the positive side:

So the statements that are true are:

The graph of the function is positive on (-co, -5).

The graph of the function is negative on (3,co).

Determine the vertex form of g(x) = x2 + 2x – 1. Which graph represents g(x)? On a coordinate plane, a parabola opens up. It goes through (negative 2, 3), has a vertex at (negative 1, 2), and goes through (0, 3). On a coordinate plane, a parabola opens up. It goes through (negative 1, 2), has a vertex at (1, negative 2), and goes through (3, 2). On a coordinate plane, a parabola opens up. It goes through (0, 3), has a vertex at (1, 2), and goes through (2, 3). On a coordinate plane, a parabola opens up. It goes through (negative 3, 2), has a vertex at (negative 1, negative 2), and goes through (1, 2).

Answers

The vertex form of the quadratic equation is:

$g(x) = (x + 1)^2 - 2$

And the graph can be seen at the end

How to get the vertex form of the quadratic equation?

For a quadratic equation with leading coefficient a, and with vertex (h, k), the vertex form is:

$y = a*(x - h)^2 + k$

Here we have:

$g(x) = x^2 + 2x - 1$

First, we need to find the vertex. by using the known formula we get:

$h = -2/(2*1) = -1$

To get the value of k, we need to evaluate g(x) in x = -1.

$g(-1) = (-1)^2 + 2*-1 - 1 = 1 - 2 - 1 = -2$

So the vertex is (-1, -2), which means that the vertex form is:

$g(x) = (x + 1)^2 - 2$

And its graph can be seen below.

If you want to learn more about quadratic equations:

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Answer:

the last one I think

Step-by-step explanation:

Prove the identity: cos(3x) + cos(x) = 2cos(2x)cos(x)I'm almost done with my homework, just got stuck on this one.

Answers

$cos \alpha +cos \beta =2\cdot cos ( \alpha + \beta )/(2) \cdot cos ( \alpha - \beta )/(2)\n-----------------\n\n \alpha =3x\ \ \ and\ \ \ \beta =x\n\n$

$L=cos \alpha +cos \beta =cos(3x)+cos(x)\n\n\Rightarrow\ \ \ 2\cdot cos ( \alpha + \beta )/(2) \cdot cos ( \alpha - \beta )/(2)=2\cdot cos (3x+x)/(2) \cdot cos (3x-x)/(2) =\n\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\cdot cos (4x)/(2) \cdot cos (2x)/(2) =\n\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\cdot cos(2x)\cdot cos (x)=R$

Make a Frequency Distribution with 5 classes18, 19, 20, 20, 21, 22, 22, 24, 25, 25, 26, 27, 28, 30, 30, 31, 33, 34, 35, 37, 37, 38, 39, 40, 41, 42, 56, 62, 73
What is the class width?
List the midpoints, relative frequency, and cumulative relative frequency
Make a relative frequency ogive
Make a frequency polygon
Calculate the mean
Calculate the median
Calculate the sample standard deviation (our data is sample from our class, not a population!)
Calculate the Q1 and Q3 values

Answers

The class width is calculated by dividing the range of the data by the number of classes. In this case, the range is 73-18=55. So, the class width is 55/5=11.

The midpoints are calculated by adding the lower and upper limits of each class and dividing by 2. The midpoints for each class are: 20.5, 31.5, 42.5, 53.5, and 64.5.

The relative frequency is calculated by dividing the frequency of each class by the total number of data points (28 in this case). The relative frequencies for each class are: 0.1071, 0.1071, 0.2143, 0.2857, and 0.2857.

The cumulative relative frequency is calculated by adding up the relative frequencies for each class and all previous classes. The cumulative relative frequencies for each class are: 0.1071, 0.2143, 0.4286, 0.7143, and 1.

To make a relative frequency ogive, you would plot the cumulative relative frequencies against the upper limits of each class.

To make a frequency polygon, you would plot the midpoints of each class against their respective frequencies.

The mean is calculated by adding up all the data points and dividing by the total number of data points (28 in this case). The mean is approximately 32.39.

The median is calculated by finding the middle value when all the data points are arranged in order from smallest to largest. Since there are an even number of data points (28), we take the average of the two middle values (25 and 30). So, the median is (25+30)/2=27.5.

The sample standard deviation is calculated using the formula: sqrt(sum((x-mean)^2)/(n-1)), where x is each data point, mean is the mean of all the data points, n is the total number of data points (28 in this case). The sample standard deviation is approximately 12.87.

The Q1 value (the first quartile) is calculated by finding the median of the lower half of the data points (14 in this case). Since there are an even number of data points in this half (14), we take the average of the two middle values (22 and 22). So, Q1 is (22+22)/2=22.

The Q3 value (the third quartile) is calculated by finding the median of the upper half of the data points (14 in this case). Since there are an even number of data points in this half (14), we take the average of the two middle values (34 and 35). So, Q3 is (34+35)/2=34.5.

This mathematical question focused on descriptive statistics, such as calculating the class widths for a frequency distribution,

Determining the midpoints, relative frequency, and cumulative relative frequency, drawing a relative frequency ogive and frequency polygon, and calculating the mean, median, sample standard deviation, Q1, and Q3.First, to construct a frequency distribution with a total of 5 classes, we need to determine the class width. We get this by subtracting the smallest value from the largest value and dividing by the number of classes, then rounding up. In this case, (73-18)/5 = 11. Therefore, the class width is 11.Next, we calculate the midpoints of each class, relative frequency, and cumulative relative frequency. After that, we create the relative frequency and frequency polygon. Unfortunately, without a greater context, these cannot be shown here.For the mean, we sum up all the numbers and divide by the number of observations. In this case, the mean is the sum of the values divided by 29.We calculate the median, which is the middle value when the numbers are arranged in ascending order. For this dataset, the median would be the 15th data value.The sample standard deviation is a little more complex. It involves finding the mean, subtracting each value from the mean and squaring the result, summing these squared values, dividing by the number of observations minus 1, and taking the square root. This gives the sample standard deviation.Lastly, Q1 and Q3 are the 25th and 75th percentiles, respectively. Q1 and Q3 can be calculated by sorting the data in ascending order and taking the 25th and 75th percentile positions.

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California has 143deaths from heart disease and 137deaths from cancer per100,000 residents. Which rate is more extreme compared to other states? A.Perform any needed computations here: b. Is California's rate for heart disease or for cancer more extreme compared to the other states? Explain:

Answers

Answer:

California's rate for heart disease is more extreme than Cancer

Step-by-step explanation:

Given

Represent Cancer with C and Heart Disease with H

H = 143

C = 137

Population = per 100,000

First we need to determine the probability of both.

For H

P(H) = H/100000

P(H) = 143/100000

P(H) = 0.00143

For C

P(C) = C/100000

P(C) = 137/100000

P(C) = 0.00137

By comparison,

0.00143 > 0.00137

So, California's rate for heart disease is more extreme than Cancer.

Reason: P(H) > P(C)

How do u check your answer for 4x+12-2x=26

Answers

Answer:

Step-by-step explanation:

4x+12-2x=26

Or, 2x=26-12

Or, 2x=14

Or, x=7

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