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Korolek [52]
2 years ago
11

An ant walks 9 3/4 feet in 3/4 minute. The distance in feet the ant walks is directly proportional to the time in minutes the an

t walks. What is the constant of proportionality that relates the distance in feet the ant walks to the time in minutes it takes the ant to walk that distance?​

Mathematics
1 answer:
OLga [1]2 years ago
4 0

Answer:

7\frac{5}{16} ANSWER (C.

Step-by-step explanation:

Multiply 9\frac{3}{4} by \frac{3}{4} to get \frac{117}{16}

-

7\frac{5}{16} = \frac{117}{16}

because 7 multiply by 16 plus 5 = 117

So 117 over 16

is the same as  7\frac{5}{16} .

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After an overpayment last month, the balance on a customer’s billing statement is −$50.00. This month’s new charges are $124.80.
siniylev [52]

The customer’s current balance after an overpayment last month is D. $174.80.

<h3>How to illustrate the information?</h3>

It should be noted that after an overpayment last month, the balance on a customer’s billing statement is −$50.00 and this month’s new charges are $124.80.

Therefore, the customer’s current balance will be:

= $124.80 - (-$50)

= $124.80 + $50

= $174.80

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3 0
1 year ago
I really need help!! can u pwease help ;-;
Nata [24]

Answer:

b

Step-by-step explanation:

6 0
3 years ago
First person to answer 10 points
Assoli18 [71]
It’s the third one. Hopefully this helped
8 0
3 years ago
Is (-4x+9) squared a perfect square
kaheart [24]

Answer:

+ 6x + 9 ...so x2 + 6x + 9 is a perfect square trinomial. ... The first term, 16x2, is the square of 4x, and the last term, 36, is the square of 6. (4x)2 – 48x + 62.

Step-by-step explanation:

5 0
3 years ago
Express the system as AX = B; then solve using matrix inverses
Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]

\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B

Adj.A=Transpose of cofactor of Matrix A

Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2

2.

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

   A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]

\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4

7 0
3 years ago
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