What is one thousand less than 89,257

The standard form equation of the circle with endpoints (-10, -8) and (-6, -2) is (x + 8)^2 + (y + 5)^2 = 13.

To find the standard form equation of a circle, we need to find the center and the radius. Since the endpoints of the diameter are given, we can find the center by finding the midpoint of the diameter. The x-coordinate of the center is the average of the x-coordinates of the endpoints, and the y-coordinate of the center is the average of the y-coordinates of the endpoints. The radius is half the length of the diameter. Once we have the center and radius, we can plug them into the standard form equation of a circle, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.

Step 1: Find the midpoint of the diameter using the formula:
(x, y) = ((x1 + x2) / 2, (y1 + y2) / 2)
(x, y) = ((-10 + -6) / 2, (-8 + -2) / 2) = (-16 / 2, -10 / 2) = (-8, -5)

Step 2: Find the radius by halving the length of the diameter:
Length of the diameter = √((x2 - x1)^2 + (y2 - y1)^2) = √((-6 - -10)^2 + (-2 - -8)^2) = √(16 + 36) = √(52)
Radius = √(52) / 2 = √(13)

Step 3: Plug the center and radius into the standard form equation of a circle:
(x + 8)^2 + (y + 5)^2 = 13

Answer with explanation:

The Data Set is→ 10,12,2,4,24,2,7,7,9

Arranging the data in ascending order

2,2,4, 7,7,9,10, 12, 24

Number of observation is odd,

So, Median

[tex]\frac{9+1}{2}=5^{th} Observation[/tex]

5th Observation = 7

[tex]Q_{1}[/tex]

=First Quartile  of data set that is from first variate to Median value

={2,2,4,7,7}

So, [tex]Q_{1}=4[/tex]

[tex]Q_{3}[/tex]

=Third Quartile of data set that is from last variate to Median value

={7,9,10,12,24}

So, [tex]Q_{3}=10[/tex]

Minimum value in data set = 2

Maximum value in data set = 24

Plotting the Box plot

Option C , is correct Box plot.

Answer:  The answer is C

Step-by-step explanation:

Cost of lectures = $7.33 per hour

Let [tex]e[/tex] the cost of the exam hours

Let [tex]w[/tex] be the cost of the workshop hours

Let [tex]l[/tex] be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

[tex]e=2w[/tex] equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

[tex]w=2l[/tex] equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

[tex]3e+24w+12l=528[/tex] equation (1)

Now, lets find the value of [tex]l[/tex]:

Step 1.  Solve for [tex]l[/tex] in equation (3)

[tex]3e+24w+12l=528[/tex]

[tex]12l=528-3e-24w[/tex] equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

[tex]12l=528-3e-24w[/tex]

[tex]12l=528-3(2w)-24w[/tex]

[tex]12l=528-6w-24w[/tex]

[tex]12l=528-30w[/tex] equation (5)

Step 3. Replace equation (2) in equation (5) and solve for [tex]l[/tex]

[tex]12l=528-30w[/tex]

[tex]12l=528-30(2l)[/tex]

[tex]12l=528-60l[/tex]

[tex]72l=528[/tex]

[tex]l=\frac{528}{72}[/tex]

[tex]l=\frac{22}{3}[/tex]

[tex]l=7.33[/tex]

Cost of lectures  = $7.33 per hour

The volume of the old silo is  1962.5 cubic feet, the volume of the new silo is 5298.75 cubic feet and about 3336.25 greater is the volume of the new silo than the old silo.

a three dimensional shape can be defined as a solid figure or an object or shape that has three dimensions—length, width, and height.

Mr. Macady has an old cylindrical grain silo on his farm that stands 25 feet high with a diameter of 10 feet.

h=25 feet and r =5 feet

Volume of cylinder = πr²h

=3.14×5²×25

=3.14×25×25

Volume =1962.5 cubic feet.

Now let us find the volume of new cylinder which has height 30 feet and diameter is 15 feet

Radius =15/2

r=7.5

Volume = 3.14× (7.5)² × 30

=3.14×56.25× 30

V=5298.75 cubic feet

Now let us find how much the new volume of silo is greater than old silo.

Difference = 5298.75 - 1962.5

=3336.25

Hence, the volume of the old silo is  1962.5 cubic feet, the volume of the new silo is 5298.75 cubic feet and about 3336.25 greater is the volume of the new silo than the old silo.

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Answer:

3

Step-by-step explanation:

Let's substitute t = 9 and u  = 11 into the expression.

3+11t−9u

=3+11(9)−9(11)

=3+99-99=3+99−99

=3

a) Numbers greater than -8 but less than -2 are represented by the double inequality [tex]\(-8 < x < -2\).[/tex]

b) Numbers less than 0 but greater than -5 are represented by the double inequality [tex]\(-5 < x < 0\).[/tex]

c) Numbers greater than -3 but less than 3 are represented by the double inequality [tex]\(-3 < x < 3\).[/tex]

To describe each set of numbers using double inequalities, we'll define the range of numbers that satisfy each condition.

a) For numbers greater than -8 but less than -2:

[tex]\[ -8 < x < -2 \][/tex]

This double inequality represents all numbers that are greater than -8 on the number line and less than -2, but not including -2 itself.

b) For numbers less than 0 but greater than -5:

[tex]\[ -5 < x < 0 \][/tex]

This double inequality represents all numbers that are less than 0 on the number line and greater than -5, but not including 0 itself.

c) For numbers greater than -3 but less than 3:

[tex]\[ -3 < x < 3 \][/tex]

This double inequality represents all numbers that are greater than -3 on the number line and less than 3, but not including 3 itself.

In summary:

a) Numbers greater than -8 but less than -2 are represented by the double inequality [tex]\(-8 < x < -2\).[/tex]

b) Numbers less than 0 but greater than -5 are represented by the double inequality [tex]\(-5 < x < 0\).[/tex]

c) Numbers greater than -3 but less than 3 are represented by the double inequality [tex]\(-3 < x < 3\).[/tex]

Answer: 21.2

hope this helped

Final answer:

The recursive definition for the given geometric sequence is an = 5an-1, with a1 = 2. Using the geometric formula to find the 10th term, we calculate the 10th term to be 3906250.

Explanation:

The given sequence is a geometric sequence, in which each term after the first is found by multiplying the previous term by a constant ratio. In this case, the ratio is obtained by dividing the second term by the first term. Therefore, the recursive formula for the given geometric sequence is an = 5 × an-1, with the first term a1 = 2.

To find the 10th term of the sequence, we use the formula: an = a1 × r(n-1), where r is the common ratio and n is the term number. Substituting the given values we get a10 = 2 × 5(10-1) = 2 × 59.

Calculating this we find that the 10th term of the sequence is 2 × 1953125 = 3906250.

Answer:

if you want to know how many males, you have to add all the numbers up :

105 + 93 + 99 + 84 + 160 + 140 + 15 + 18 = 714

and then you divide 714 by all the male numbers 105 + 99 + 160 + 15 = 379

714/379 =  53.08 %

~batmans wife dun dun dun....

The number of wildflowers is doubling each week. Starting with 88 blooms in the first week and 176 in the second, the pattern suggests that there will be 2816 flowers after six weeks.

The student is asking about a pattern of growth in the population of wildflowers. The flowers began blooming with 88 blooms and by the second week, there were 176 flowers in bloom. This indicates that the number of flowers is doubling each week. To find the number of flowers after six weeks, we can establish a pattern of growth and calculate the number for the sixth week.

We can see that from week 1 to week 2, the blooms doubled (88 x 2 = 176). If this pattern continues, the blooms will double each week.

Let's calculate the number of blooms for each week up to six weeks:

Week 1: 88 flowers

Week 2: 176 flowers (88 x 2)

Week 3: 352 flowers (176 x 2)

Week 4: 704 flowers (352 x 2)

Week 5: 1408 flowers (704 x 2)

Week 6: 2816 flowers (1408 x 2)

The answer is B, as they both add up to 15 feet and 4 inches

[tex]\dfrac{6}{59}=\dfrac{3}{x}\qquad\text{cross multiply}\\\\(6)(x)=(3)(59)\\\\6x=177\qquad\text{divide both sides by 6}\\\\\boxed{x=29.5}[/tex]

The 6th term is ( 4 / 27 ) and the recursive formula is q = 1/3.

When there is a constant between the two successive numbers in the series then it is called a geometric series.

The 6th term of the series is calculated as:-

1st term = 36

2nd term = 12

3rd term = 4

The recursive formula is,

q = 12/36 = 1/3

The 6th term is,

[tex]a_n = a_1( r )^{n-1}[/tex]

[tex]a_6=36 ( \dfrac{1}{3})^5=\dfrac{4}{27}[/tex]

Therefore, the 6th term is ( 4 / 27 ) and the recursive formula is q = 1/3.

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The correct answer is 58 viable solutions.

To model the given situation, we need to define variables, create inequalities, and then solve the system of inequalities.

Let x = the number of pens, and y = the number of pencils.

Given information:

- The total number of writing utensils must be less than or equal to 25.

- The number of pencils must be greater than or equal to the square of (3 less than the number of pens).

Step 1: Create the first inequality for the total number of writing utensils.

x + y ≤ 25

Step 2: Create the second inequality for the condition on the number of pencils.

y ≥ (x - 3)^2

Step 3: Solve the system of inequalities.

We can substitute different values of x into the second inequality to find the corresponding values of y.

For x = 1, y ≥ (-2)^2 = 4

For x = 2, y ≥ (-1)^2 = 1

For x = 3, y ≥ 0^2 = 0

For x = 4, y ≥ 1^2 = 1

For x = 5, y ≥ 2^2 = 4

...

Substituting these values into the first inequality, we get:

1 + y ≤ 25 (y ≥ 4)

2 + y ≤ 25 (y ≥ 1)

3 + y ≤ 25 (y ≥ 0)

4 + y ≤ 25 (y ≥ 1)

5 + y ≤ 25 (y ≥ 4)

...

Therefore, the viable solutions are:

(x, y) = (3, 0), (3, 1), (3, 2), ..., (3, 22)

(x, y) = (4, 1), (4, 2), ..., (4, 21)

(x, y) = (5, 4), (5, 5), ..., (5, 20)

There are a total of 22 + 20 + 16 = 58 viable solutions.