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STatiana [176]
3 years ago
9

Compare the values of each of the following pairs of fractions by expanding the fractions

Mathematics
1 answer:
sattari [20]3 years ago
3 0

We'll need to find and use the LCD here.  The LCD is 7(9), or 63.  Then this set of fractions becomes

{81/7, 77/7}.  The former (first) fraction is the larger one:  81/7 > 77/7, so

9/7 > 11/9.

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Janelle went to the bank twice and withdrew a total of $40 from
Oduvanchick [21]

Answer:

20$ each visit

Step-by-step explanation

Divide 40 by 2 for her 2 visits to the bank

\frac{40}{2} = 20 \\

<em>Final Answer: </em>20

 

6 0
2 years ago
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Ann [662]

Answer:

The third one(-20+(n-1)*)

Step-by-step explanation:

It's correct because when you plug in one of the numbers in the list, it makes the equation equal. Example Below.

n=4

n=-20+(n-1)8

4=-20+(4-1)8

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7 0
3 years ago
Will give brainliest please help:)<br><br>Simplify: <br>-3-8(-5x - 1)​
alex41 [277]

Answer:

5+40x

Step-by-step explanation:

I hope this helps. Best of luck.

7 0
3 years ago
Read 2 more answers
The table shows the viscosity of an oil as a function of temperature. Identify a quadratic model for the viscosity, given the te
777dan777 [17]

Answer:

The viscosity at 140°C is predicted to be 7.2

Step-by-step explanation:

The function that model the relationship between viscosity and temperature = Quadratic model

The general form of a quadratic equation is y = a·x² + b·x + c

Therefore, we have;

When y = 10.8, x = 110, which gives;

10.8 = a·110² + b·110 + c = 12100·a + 110·b + c

10.8 = 12100·a + 110·b + c

When y =8.2, x = 130

8.2 = a· 130² + b· 130 + c = 16900·a + 130·b + c

8.2 = 16900·a + 130·b + c

When y = 160, x = 5.8

5.8 = a·160² + b·160 + c = 25600·a + 160·b + c

5.8 = 25600·a + 160·b + c

The three equations above can be listed as follows;

10.8 = 12100·a + 110·b + c

8.2 = 16900·a + 130·b + c

5.8 = 25600·a + 160·b + c

Solving using matrices gives;

\begin{bmatrix}12100 & 110 & 1\\ 16900 & 130 & 1\\ 25600 & 160 & 1\end{bmatrix} \begin{bmatrix}a\\ b\\ c\end{bmatrix} = \begin{bmatrix}10.8\\ 8.2\\ 5.8\end{bmatrix}

\begin{bmatrix}a\\ b\\ c\end{bmatrix} = -\dfrac{1}{3000}\begin{bmatrix}3 & -5 & 2\\ -867 & 1350 & -480\\ 62400 & -88000 & 28600\end{vmatrix} \begin{bmatrix}10.8\\ 8.2\\ 5.8\end{bmatrix}

From which we have;

a = 0.001, b = -0.37, c = 39.4

Substituting gives;

y = 0.001·x² - 0.37·x + 39.4

When x = 140

y = 0.962·140² - 0.37·140 + 39.4= 7.2

The viscosity at 140°C = 7.2.

8 0
3 years ago
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cricket20 [7]

the answer is D (i think i'm sorry if i get u wrong)

3 0
3 years ago
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