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Arturiano [62]
3 years ago
8

TerminatingRepeatingDecimalsCardSortActivity

Mathematics
1 answer:
yulyashka [42]3 years ago
8 0
Is there supposed to be a picture or is this just random
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Use the distributive property to remove the parentheses, simplify.
Nookie1986 [14]
Hi again :)

5b^3(4 - 9b^2)

Use the distributive property

=(5b^3)(4) + (5b^3)(-9b^2)

= 20b^3 - 45b^5

= -45b^5 + 20b^3

Have a nice day!
8 0
3 years ago
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Determine the relation of the function.
Arte-miy333 [17]
The last one. to be a function you cannot have 2 points on the same vertical line
8 0
3 years ago
A salesperson's weekly paycheck is 10% more than a second salesperson's paycheck. The two paychecks total $1525. Find the amount
Kaylis [27]
1525 ➗ 2 = 762.5
762.50 x 10% is 76.25
762.50 - 76.25 = 686.25
one paycheck is 762.50
the other is 686.25

I hope this helps.
6 0
3 years ago
The number of students in the tutoring center was recorded for 47 randomly selected times. The data is summarized in the frequen
ddd [48]

Using the information given, it is found that the class width for this frequency distribution table is of 1.

In this problem, these following classes are given:

0 – 1 14

2 – 3 1

4 – 5 8

6 – 7 12

8 – 9 12

The classes not given, which are 1 - 2, 3 - 4 and 5 - 6, have values of 0.

The <u>difference between the bounds of the classes is of 1</u>, thus, the class width is of 1.

A similar problem is given at brainly.com/question/24701109

5 0
2 years ago
Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.
skad [1K]

Answer:

The absolute maximum and minimum is 20\; \text{and} -20.

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

f_x=y-4=0

f_y=x=0

The only solution to this system of equations is the point (0, 4), which lies in the domain.

f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1

\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1

\therefore (0,4) is a saddle point.

Boundary points -  (4,0),  (-4,0), (0,16)

Along boundary  y=16-x^2

   f=x(16-x^2)-4x

=16x-x^3-4x

\Rightarrow f^'=16-3x^2-4=0

\Rightarrow 3x^2=12

\Rightarrow x=\pm2,\;\;y=14

Values of f(x) at these points.

\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}

Therefore, the absolute maximum and minimum is 20\; \text{and} -20.

6 0
3 years ago
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