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2 years ago

From the sum of 3a^2-ab-2b^2 and 2a^2+5ab-3b^2 subtract a^2-3ab-4b^2

Mathematics

2 answers:

melamori03 [73]2 years ago

4 0

Answer:

4a^2 + ab - 9b^2

Step-by-step explanation:

First, perform the indicated addition:

3a^2-ab-2b^2

2a^2+5ab-3b^2

------------------------

5a^2 + 4ab - 5b^2

From this sum we subtract a^2 - 3ab - 4b^2:

5a^2 + 4ab - 5b^2

-(a^2 - 3ab - 4b^2

------------------------------

4a^2 + ab - 9b^2

Luden [163]2 years ago

3 0

4a^2+7ab-b^2

Step-by-step explanation:

Add the first two equations together:

3a^2-ab-2b^2

+ 2a^2+5ab-3b^2

---------------------------

5a^2+4ab-5b^2

Subtract that answer from the remaining trinomial:

5a^2+4ab-5b^2

- a^2-3ab-4b^2

--------------------------

4a^2+7ab-b^2

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3 years ago

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NISA [10]

Answer:

Class second have higher score and have greater spread.

Step-by-step explanation:

For first box plot

$\text{Minimum value }= 53$

$Q_1=62$

$Median=80$

$Q_3=86$

$\text{Maximum value }= 89$

$Range=Maximum-Minimum=89-53=36$

$IQR=Q_3-Q-1=86-62=24$

For second box plot

$\text{Minimum value }= 56$

$Q_1=62$

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$\text{Maximum value }= 96$

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$IQR=Q_3-Q-1=89-62=27$

First class has greater minimum value, it means first class has lower grades.

First quartile of both classes are same, it means equal number students in both classes have less than 62 marks.

First class has greater median.

second class has greater third quartile.

Second class has greater Maximum value. It means second class have higher score than first class.

Second class has greater range. It means the data of second class has greater spread.

Second class has greater inter quartile range. It means the data of second class has greater spread.

Therefore, the class second have higher score and have greater spread.

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3 years ago

Recipe for a cake needs 1 1/2 cups of milk. You are making 1/2 of the recipe. How much milk do you need

nalin [4]

Answer:

3/4

Step-by-step explanation:

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3 years ago

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devlian [24]

Answer:

h = 17.65 m

Step-by-step explanation:

The given equation gives the speed at impact :

$v=\sqrt{2gh}$

h is height form where the object is dropped

Put v = 18.6 m/s in the above equation.

$h=\dfrac{v^2}{2g}\\\\h=\dfrac{(18.6)^2}{2\times 9.8}\\\\h=17.65\ m$

So, the object must be dropped from a height of 17.65 m.

4 0

3 years ago