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zheka24 [161]
3 years ago
13

Dustin has three lengths of rope: 3/4 yard long, 2/3 yard long, and 7/10 yard long. He needs at least 1 yard of rope. Which two

pieces of rope together will be at least 1 yard in length and have the smallest amount left over?
Mathematics
2 answers:
nordsb [41]3 years ago
7 0
Fisrt you have to find the common denominator, 60. 40/60, 45/60, and 42/60. the answer is 3/4 and 7/10
ad-work [718]3 years ago
6 0
A think it is all of the above because each rope is the same amount
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The number of lines of reflection about which the combined figure can reflect onto itself is ____
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A cereal box has a height of 32 cm. The base has 160 square cm what is the volume in cubic cm of the box
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Automobile license plates in Massachusetts usually consist of three digits followed by three letters. The first digit is never z
Serga [27]

Answer:

1,58,18,400

Step-by-step explanation:

1st digit can have the values 1-9 (9 distinct values)

2nd digit can have the values 0-9 (10 distinct values)

3rd digit can have the values 0-9 (10 distinct values)

1st letter can have the value A-Z (26 distinct values)

2nd letter can have the value A-Z (26 distinct values)

3rd letter can have the value A-Z (26 distinct values)

Total number of different plates possible = 9*10*10*26*26*26

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3 0
3 years ago
Solve the system by using a matrix equation (Picture provided)
Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

10x - 6y =2

it can be represented in matrix form as\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]

A= \left[\begin{array}{cc}7&-4\\10&-6\end{array}\right]

X= \left[\begin{array}{c}x\\y\end{array}\right]

B= \left[\begin{array}{c}4\\2\end{array}\right]

i.e, AX=B

or X= A⁻¹ B

A⁻¹ = 1/|A| * Adj A

determinant of A = |A|= (7*-6) - (-4*10)

                                    = (-42)-(-40)

                                    = (-42) + 40 = -2

so, |A| = -2

Adj A=  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]

A⁻¹ =  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]/ -2

A⁻¹ =  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]

X= A⁻¹ B

X=  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]

X= \left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]

X= \left[\begin{array}{c}12-4\\20-7\end{array}\right]

X= \left[\begin{array}{c}8\\13\end{array}\right]

x= 8, y= 13

solution set= (8,13).

Option b is correct.

3 0
3 years ago
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