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She needs 450 so 450-(95+24)

                              450-(119)=331

331/5=66.2 but, you need round up another jar to make it a whole number 

In conclusion, the solution to this question is 67.

Let's denote:

P = number of professors

L = number of lecturers

We know:

P + L = 228 (total faculty members)

P = 5 * L (5 professors for every 14 lecturers)

Substituting the second equation into the first:

5 * L + L = 228

6 * L = 228

L = 228 / 6

L = 38

Now, we can find the number of professors:

P = 5 * L

P = 5 * 38

P = 190

Therefore, the community college employs:

190 professors

38 lecturers

In total, they employ 228 faculty members.

Find the smallest positive integer B so that when we express the decimal number 301 as a base B number, we still get a 2-digit number.   

Well, in base 10 it's a three digit number, so the base has to be bigger than ten.  

Per https://www.rapidtables.com/convert/number/base-converter.html it's base 18 ... 301₁₀ = GD₁₉    

For information, the next one down was 17 ... 310₁₀ = 10C₁₇    

_.   

If we multiply both sides by 3, we get \(3x + 3y = 75\) . We notice that \(6x+3y\) is \(3x\) more than \(3x + 3y\). If \(x\) is 0, then \(3x\) is 0. 75 + 0 = 75, so the minumum value of \(6x + 3y\) is 75.

we get 3x+3y=75. since we need to add 3x, we just make x=0, since the questions says "nonnegative". therefore, the minimum value is 75.

so, we can simplify the equation to be \(4x^2-8x+4y^2+14y=-12.\)

dividing the equation by 4, we get \(x^2-2x+y^2+7y/2=-3\)

completing the square, we get \((x-1)^2+(y+\frac{7}{4})^2=\frac{17}{16}\)

17/16 is the radius squared, so we take the square root and get \(\boxed{\frac{\sqrt{17}}{4}}\)

well, since they paint 1000 sq ft in 10 minutes, the individual work rate is 10 sq ft per minute.

plug that into the second equation, and you get 5(members)*10(work rate)*x(time)=1000. so x=20, 20 minutes.

I hope this is right, it seems simple enough .

Understanding the Problem:

We have triangle ABE with points C and D on side BE.

We know the lengths of BD and CE, as well as the areas of triangles ABC and ADE.

We need to find the area of triangle ACD.

Strategy:

Find the ratio of areas of triangles ABC and ADE.

Use this ratio to find the ratio of heights from A to BC and DE.

Use this ratio to find the ratio of bases BC and DE.

Use the given lengths of BD and CE to find the lengths of BC and DE.

Use the lengths of BC, DE, and the ratio of their areas to find the area of triangle ACD.

Solution:

Ratio of areas:

Area of ABC / Area of ADE = 10 / 24 = 5 / 12

Ratio of heights:

Since the triangles have the same base (AE), the ratio of their areas is equal to the ratio of their heights from A.

Height of ABC / Height of ADE = 5 / 12

Ratio of bases:

Since the triangles have the same height from A, the ratio of their areas is equal to the ratio of their bases.

BC / DE = 5 / 12

Lengths of BC and DE:

BD = 8 and CE = 12

BC = BD + DE = 8 + DE

Substituting BC in the ratio: (8 + DE) / DE = 5 / 12

Solving for DE: DE = 48/7

BC = 8 + 48/7 = 104/7

Area of ACD:

Area of ACD / Area of ADE = BC / DE = 5 / 12

Area of ACD = (5/12) * 24 = 10

Therefore, the area of triangle ACD is 10 square units.

To determine the area of a square inscribed in a right triangle with legs of length 1 and 3, we can use a geometric approach.

Let's denote the right triangle \(ABC\) where \(A = (0, 0)\), \(B = (1, 0)\), and \(C = (0, 3)\). The right angle is at \(A\), the x-axis extends along \(AB\), and the y-axis along \(AC\).

Let the side length of the inscribed square be \(s\). When a square is inscribed in the triangle, the square's lower left corner will coincide with point \(A\) (the right angle), and the top right corner will lie somewhere along the hypotenuse \(BC\).

We need to find the relationship between \(s\) and the coordinates of the square touching the hypotenuse. The coordinates where the square touches the hypotenuse will be at the point \((s, s)\) since the square's sides are parallel to the axes.

Now, we need to find the equation of line \(BC\). The slope \(m\) of line segment \(BC\) can be calculated as follows:

\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - 1} = -3.
\]

The equation of line \(BC\), which has a y-intercept at point \(C\) \((0, 3)\), can be expressed as:

\[
y - 3 = -3(x - 0) \quad \Rightarrow \quad y = -3x + 3.
\]

Now, we want the point \((s, s)\) on the line \(BC\). Therefore, substituting \(x = s\) into the line equation gives:

\[
s = -3s + 3.
\]

Solving this equation for \(s\):

\[
s + 3s = 3 \quad \Rightarrow \quad 4s = 3 \quad \Rightarrow \quad s = \frac{3}{4}.
\]

The area \(A\) of the square is given by \(s^2\):

\[
A = s^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}.
\]

Thus, the area of the square inscribed in the right triangle is

\[
\boxed{\frac{9}{16}}.
\]

First, let's calculate the distance Will travels before grace even leaves. 

Since he traveled for 45 minutes, we have

\((45 min)(50 m/min) = 2250 m\)

So when Grace starts, she and Will have

\((2800 m) – (2250 m) = 550 m\) to cover. 

Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing. 

We have the equation

\((50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875\)

This is approximately 6 minutes and 52 seconds. 

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

Thus, our final answer is \(2:51:52\)

Thanks! :)

We can complete this problem in two different ways. 

The first tactic is to essentially compare this root to the quadratic equation of \(5x^2+21x+v = 0\)

From this quadratic, we can identify that a = 5, b = 21, c = v. 

We have the equation

\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\\ \frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\\ \sqrt{201} = \sqrt{441-20v}\\ 201 = 441-20v\\ v=12\)

This is a bit complicated and takes a lot of computations, but it does give us the correct answer. 

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The second tactic is to use conjugations of square roots. 

This is because the conjugate root theorem states that if a root of a polynomial is a square root \(a+\sqrt b\), then its conjugate, \(a-\sqrt b\) is also a root

We can apply that to this problem. If \(\frac{-21-\sqrt{201}}{10}\)  is a root, then \(\frac{-21 + \sqrt {201}}{ 10 }\) is also a root. 

The product of the roots is \([ (-21)^2 - 201 ] / 100 = 240/100 = 2.4\)

However, in the quadratic, we also have that \(v/5\) is also equal

Thus, we have 

\(v/5 = 2.4\\ v = 12\)

SO 12 is the final answer. 

Thanks! :)