The number of boys varies directly as the number of girls and inversely as the number of teachers when there are 24 boys and 16 girls there are 2 teachers how many girls are there when there are 73 boys and 1 teacher?

Answer:

B.) 2 lb 2 oz

Step-by-step explanation:

$1.79 for 20 ounces gives us

1.79/20 = $0.0895 ≈ $0.09 per ounce.

1 lb = 16 oz.; this means that 2 lb = 2(16) = 32 oz.

$2.59 for 2 lb 2 oz then gives us

2.59/34 = $0.07617647059 ≈ $0.08 per ounce.

The second choice is cheaper by 1 cent.

Answer: The correct options are (b) and (d).

Explanation:

It the polar form [tex]r^2=x^2+y^2[/tex], where

[tex]x=r\cos \theta,y=r\sin \theta[/tex]

The polar coordinate are in the form of [tex](r,\theta)[/tex].

From the given figure it is noticed that the value of r is 4 and [tex]\theta=\frac{\pi}{2}[/tex] or [tex]90^{\circ}[/tex] .

So the point is defined as [tex](4,90^{\circ})[/tex] and option b is correct.

The value,

[tex](r\cos \theta, r\sin \theta)=(0,4)[/tex]

Check the each option if we get the same value then that option is correct.

For option a.

[tex](r\cos \theta, r\sin \theta)=(-4\cos 90^{\circ} , -4\sin 90^{\circ})=(0,-4)[/tex]

Therefore option (a) is incorrect.

For option c.

[tex](r\cos \theta, r\sin \theta)=(4\cos (-90)^{\circ} , 4\sin (-90)^{\circ})=(0,-4)[/tex]

Therefore option (c) is incorrect.

For option d.

[tex](r\cos \theta, r\sin \theta)=(-4\cos (270)^{\circ} , -4\sin (270)^{\circ})\\(-4\cos (360-90)^{\circ} , -4\sin (360-90)^{\circ})=(0,4)[/tex]

Therefore option (d) is correct.

For option (e).

[tex](r\cos \theta, r\sin \theta)=(-4\cos (-270)^{\circ} , -4\sin (-270)^{\circ})\\(-4\cos (270)^{\circ} , 4\sin (270)^{\circ})=(0,-4)[/tex]

Therefore option (e) is incorrect.

Answer:

1. A

2. B, D

3. A, D, E

4. C

5. A

Step-by-step explanation:

The rates charged per hour by the first and second mechanics will be  $105 and $50.

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

It is given that two mechanics worked on a car, the first mechanic worked for 10 hours, and the second worked for 15 hours. together they charged a total of 1800.

Suppose x is the hourly rate for the first mechanic and y is the hourly rate of the second.

If the total money earned is 1800.

10x + 15 y = 1800  ---- (1)

If both have their individual rates the sum of the hourly rate is,

x+ y = 15            -------- (2)

Rearrange the equation,

x = 155- y

Substitute the value of x in equation 1 we get,

10x + 15 y = 1800

10(155-y) + 15y = 1800

1150 - 10y + 15y = 1800

5y = 250

y = 50

Substitute the value of y in equation 2 we get,

x = 155- y

x= 155 - 50

x = 105

Thus, the rates charged per hour by the first and second mechanics will be  $105 and $50.

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[tex]\[\{(3 + \sqrt{5}, 3 - \sqrt{5})}\][/tex] is the one pair of real numbers [tex]\((x, y)\)[/tex] that satisfies the given equations.

Given the system of equations:

[tex]\[x + y = 6\][/tex]

[tex]\[x^3 + y^3 = 144\][/tex]

We can use the identity for the sum of cubes:

[tex]\[x^3 + y^3 = (x + y)(x^2 - xy + y^2)\][/tex]

Substitute [tex]\(x + y = 6\):[/tex]

[tex]\[x^3 + y^3 = 6(x^2 - xy + y^2)\][/tex]

Given [tex]\(x^3 + y^3 = 144\)[/tex], we substitute this into the equation:

[tex]\[144 = 6(x^2 - xy + y^2)\][/tex]

Divide both sides by 6:

[tex]\[24 = x^2 - xy + y^2\][/tex]

Next, we use the square of the sum:

[tex]\[(x + y)^2 = x^2 + 2xy + y^2\][/tex]

Substitute [tex]\(x + y = 6\):[/tex]

[tex]\[6^2 = x^2 + 2xy + y^2\][/tex]

[tex]\[36 = x^2 + 2xy + y^2\][/tex]

We already have [tex]\(x^2 - xy + y^2 = 24\)[/tex]. Let’s denote the first and second equations:

1. [tex]\(x^2 - xy + y^2 = 24\)[/tex]

2. [tex]\(x^2 + 2xy + y^2 = 36\)[/tex]

Subtract the first equation from the second equation:

[tex]\[(x^2 + 2xy + y^2) - (x^2 - xy + y^2) = 36 - 24\][/tex]

[tex]\[3xy = 12\][/tex]

[tex]\[xy = 4\][/tex]

Now, we have two equations:

1. [tex]\(x + y = 6\)[/tex]

2. [tex]\(xy = 4\)[/tex]

We solve these equations using the quadratic formula. Consider the quadratic equation:

[tex]\[t^2 - (x+y)t + xy = 0\][/tex]

[tex]\[t^2 - 6t + 4 = 0\][/tex]

Solve for [tex]\(t\)[/tex] using the quadratic formula [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):[/tex]

[tex]\[t = \frac{6 \pm \sqrt{36 - 16}}{2}\][/tex]

[tex]\[t = \frac{6 \pm \sqrt{20}}{2}\][/tex]

[tex]\[t = \frac{6 \pm 2\sqrt{5}}{2}\][/tex]

[tex]\[t = 3 \pm \sqrt{5}\][/tex]

Thus, the solutions for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are:

[tex]\[x = 3 + \sqrt{5}, \quad y = 3 - \sqrt{5}\][/tex]

or

[tex]\[x = 3 - \sqrt{5}, \quad y = 3 + \sqrt{5}\][/tex]

Answer:

422 in.2

Step-by-step explanation:

Substitute the measurements of the rectangular prism into the formula SA = 2(lw) + 2 (lw) + 2(lh): 2(13)(7) + 2(6)(7) + 2(13)(6) = 422 in2.

Answer:

[tex]x=\frac{5}{(10-a-b)}[/tex]

Step-by-step explanation:

4+3x-ax=9-7x+bx

take all variable terms at one side and all constant terms at one side,

3x+7x-bx-ax=9-4

10x-ax-bx=5

x(10-a-b)=5

[tex]x=\frac{5}{(10-a-b)}[/tex]

The inequality the graph represent is y > x - 2.

Inequality is defined as the relation which makes a non-equal comparison between two given functions. An inequality is a mathematical statement that compares two values or expressions using a relational operator, such as less than (<), greater than (>), less than or equal to (≤), greater than or equal to (≥), or not equal to (≠). Inequalities are used to describe relationships between quantities or to express constraints or conditions.

We are given that;

The number line

Now,

The slope of the line is 1 and the y-intercept is -2,

so we can write y = x - 2 for the equation of the line.

The line is dashed, so we know that y is not equal to x - 2.

The region above the line is shaded, so we know that y is greater than x - 2.

Therefore, by the inequality the answer will be y > x - 2.

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Answer: f(x) > g(x).

If f(x) = 1000(3)x and g(x)=8x

fx=3000x

f=3000x/x

f=3000

g(x)=8x

g=8x/x

g=8

For whatever amount of X (positive), the statement f(x) > g(x) is true.

The ship's speed in deep water is 0.045 nautical miles per minute. In shallow water, the ship's maximum speed typically decreases due to increased drag and turbulence.

To find the speed of the ship in very deep water:

Answer:

Ruth was out of Portland at 35mph at 7:00 am

so at 9 am, she was 70 miles ahead of Michael

but Michael was 20 miles ahead by 3 pm.

so in 6 hours, he went past Ruth by 90 miles which is 90/6 = 15mph

Michael speed = ruth speed + 15. = 35 + 15 = 50mph

Here is your answer:

All you need to do is keep the 29 and then add a hundred because of how many 9's there are.

Also, you could see that:

[tex] .29999999999 \neq \frac{29999999999}{100000000000} [/tex]

Which would also mean that:

[tex] \frac{29999999999}{100000000000} \neq \frac{29}{100} [/tex]

Therefore the answer is "29/100."

Hope this helps!

Answer:

13/46

Step-by-step explanation:

The number of balls in 7th line would be 10.

Given is a pattern as shown in the image.

       a, a + d, a + 2d, a + 3d, ......,

We can write the pattern as -

4, 5, 6, 7, .....

This is a arithmetic progression. We can write the nth term as -

a{n} = a + (n - 1)d

a{7} = 4 + (7 - 1) x 1

a{7} = 4 + 6

a{7} = 10

Therefore, the number of balls in 7th line would be 10.

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The error interval for a number x rounded to 12.3 is 12.25 ≤ x < 12.35, indicating the range that x could be before rounding to one decimal place.

If a number x, rounded to 1 decimal place, is 12.3, the error interval for x defines the range of values that x could be in order to round to 12.3. Since the value is rounded to one decimal place, we consider half of a tenth (0.05) on either side of 12.3 to find the error interval. Hence, the lower bound of the error interval would be just below 12.3 (12.25), and the upper bound would be just before it would round up to the next tenth (12.35).

The error interval for x can be written as: 12.25 ≤ x < 12.35.

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Error interval for [tex]\( x \)[/tex] rounded to 1 decimal place as 12.3 is from 12.25 to 12.35.

To find the error interval for the number [tex]\( x[/tex], rounded to 1 decimal place as 12.3, we need to consider the possible range of values that [tex]\( x \)[/tex] could have been before rounding.

1. Lower Bound: To find the lower bound, we need to consider the smallest value [tex]\( x \)[/tex] could have been that would round to 12.3 when rounded to 1 decimal place. The smallest value that would round up to 12.3 is 12.25.

2. Upper Bound: Similarly, to find the upper bound, we need to consider the largest value [tex]\( x \)[/tex] could have been that would round to 12.3 when rounded to 1 decimal place. The largest value that would round down to 12.3 is 12.35.

So, the error interval for [tex]\( x \)[/tex] is from 12.25 to 12.35.

The measure of a complementary angle to a 27 degree angle is 63 degrees because two angles are complementary if their measures add up to 90 degrees. This is calculated by subtracting the given angle from 90.

The measure of an angle is given as 27°, and you are asked to find the measure of a complementary angle. In mathematics, two angles are considered complementary if their measures add up to 90 degrees. Therefore, to find the measure of a complementary angle, you subtract the given angle from 90 degrees.

So, to solve this problem, you would simply subtract 27 from 90.
The calculation would look like this:
90-27 = 63. Therefore, the measure of the complementary angle is 63°.

Answer:

C. Either 46.27 miles or 17.43 miles

The radius of the circle that includes the footbridge, given the bridge's arc length and height, is 14.1 ft.

The problem relates to the geometry of a circle, specifically the concept of the arc length and radius. Given the arc length (L) and the height of the arc (h), we can use these to find the radius (r) of the circle using the formula r = L²/(8h) + h/2.

Here, the arc length is the horizontal length of the footbridge, which is 27 ft and the height h is the height of the bridge, 10 ft.

Substituting these values into the formula we get:

r = (27² ft)/(8*10 ft) + 10 ft/2 = 729/80 + 5  = 14.1 ft

So, to the nearest tenth of a foot, the radius of the circle that includes the footbridge is 14.1 ft.

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