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Luden [163]
3 years ago
10

Casey's drill team squad was responsible for painting props for a performance at a band contest. The squad painted 18 props in 9

0 minutes. If they continued painting at this rate, how many props would they paint in 3 hours?
Mathematics
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer: They would paint 36 props in 3 hours.

Step-by-step explanation:

Given: The squad painted 18 props in 90 minutes.

Time taken to paint each prop = \dfrac{90\text{ minutes}}{18}

= 5 minutes

Now, 1 hour = 60 minutes

3 hours = 3 x 60 minute = 180 minutes

Total props painted in 3 hours = (180 )÷5

= 36

Hence, with same rate , they would paint 36 props in 3 hours.

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HELP
Shtirlitz [24]

x(x + 9-x) = 6^2

9x = 36

x = 4


Answer

c. 4

4 0
3 years ago
The difference between the roots of the quadratic equation x2+x+c=0 is 6. Find c.
NemiM [27]
We have that
x_{1} - x_{2} = 6
Let y be the first root, then y - 6 will be the second one. Since both roots make the quadratic equation equal to 0, we have that:
y^{2}  + y + c =  (y - 6)^{2}  + (y - 6) +c
Solve for y:
y^{2} +y +c = y^{2} - 12y + 36 + y - 6 +c \\ y + 12y - y = 30 \\ y =  \frac{5}{2}
Plug in y = 5/2 into original quadratic equation and solve for c:
(  \frac{5}{2}) ^{2} +  \frac{5}{2}  +c = 0 \\  c= - \frac{5}{2}  -  \frac{25}{4}  \\ c = - \frac{35}{4}
So, the answer is c = -35/4
6 0
4 years ago
StephanieHad 7/8 pound of bird seed she can use 3/8 pound to fill a bird feeder how much birdseed to Stephanie have left
Anna11 [10]
7/8 - 3/8= 4/8 I hope this helps.
4 0
4 years ago
Write as a single fraction in its simplest form.
yaroslaw [1]

\dfrac{4}{3x-5} + \dfrac{x+2}{x-1}\\\\\\=\dfrac{4(x-1) + (x+2)(3x-5)}{(3x-5)(x-1)}\\\\\\=\dfrac{4x-4 + 3x^2 -5x +6x -10}{(3x-5)(x-1)}\\\\\\=\dfrac{3x^2 +5x-14}{(3x-5)(x-1)}

5 0
2 years ago
Find the reduced row echelon form of the following matrices and then give the solution to the system that is represented by the
GaryK [48]

Answer:

a)

Reduced Row Echelon:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

Solution to the system:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

Reduced Row Echelon:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

Solution to the system:  

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

Step-by-step explanation:

To find the reduced row echelon form of the matrices, let's use the Gaussian-Jordan elimination process, which consists of taking the matrix and performing a series of row operations. For notation, R_i will be the transformed column, and r_i the unchanged one.

a) \left[\begin{array}{cccc}0&4&7&0\\2&1&0&0\\0&3&1&-4\end{array}\right]

Step by step operations:

1. Reorder the rows, interchange Row 1 with Row 2, then apply the next operations on the new rows:

R_1=\frac{1}{2}r_1\\R_2=\frac{1}{4}r_2

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right]

2. Set the first row to 1

R_3=-3r_2+r_3

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

3. Write the system of equations:

x_1+\frac{1}{2}x_2=0\\x_2+\frac{7}{4}x_3=0\\x_3=-4

Now you have the  reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right]

1. R_2=-2r_1+r_2\\R_3=-r_1+r_3

Resulting matrix:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

2. Write the system of equations:

4x_1+3x_2=7\\2x_3=-17

Now you have the reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

7 0
3 years ago
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