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kolezko [41]
3 years ago
8

What is 3/5 of $40 restaurant bill

Mathematics
1 answer:
iVinArrow [24]3 years ago
6 0
24 is the answer ,multiply 3\5 by 40
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Crackers were on sale 2 for 5.50. I bought three packs, How much did I spend?
MrRissso [65]

Answer:

6.75

Step-by-step explanation:

2 crackers=5.50

1 cracker=5.50÷2

=2.25

3 crackers=2.25×3

=6.75

8 0
2 years ago
Read 3 more answers
Use matrices to solve the system of equations if possible. Use Gaussian elimination with back substitution or gauss Jordan elimi
CaHeK987 [17]

In matrix form, the system is given by

\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}

I'll use G-J elimination. Consider the augmented matrix

\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]

• Multiply through row 1 by -1.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]

• Multiply row 3 by 1/3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]

Then the solution to the system is

\boxed{x=9, y=-3, z=8}

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]

The third row tells us that z=8. Then in the second row,

y-z = -11 \implies y=-11 + 8 = -3

and in the first row,

x-y+z=20 \implies x=20 + (-3) - 8 = 9

5 0
2 years ago
Use the zeros and the labeled point to write the quadratic function
Anarel [89]

Answer:

y = 2x2 - 10x + 12

Step-by-step explanation:

5 0
2 years ago
If VT=2,what is the length of PT?
marusya05 [52]

Step-by-step explanation:

\underline{ \underline{ \text{Given : }}}

  • Perpendicular ( P ) = VT = 2
  • Base ( b ) = PT
  • \theta =  \tt{60 \degree}

\underline{ \underline{ \text{Solution}}} :

\tt{ \tan(60 \degree)  =  \frac{perpendicualar}{base}}

⟶ \tt{ \sqrt{3}  =  \frac{2}{PT}}

⟶ \tt{ \sqrt{3}  \: PT= 2}

⟶ \tt{PT =  \frac{2}{ \sqrt{3}} }

⟶ \tt{PT= \boxed{\tt{ \frac{2 \sqrt{3} }{{3} }}}}

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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6 0
3 years ago
You work at a craft store that sells paint kits. The kits have 4 thin paintbrushes for every 6 thick brushes. What is
liq [111]

Answer:2:3

Step-by-step explanation:

if you were to simplify the ratio just like you would a fraction you would get 2:3

6 0
3 years ago
Read 2 more answers
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