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3 years ago

SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

Mathematics

2 answers:

lapo4ka [179]3 years ago

7 0

Answer:

-4a^2 -7a^2 - 2ab +3ab +9b^2 -5b^2

11a^2 + ab +4b^2

Step-by-step explanation:

LekaFEV [45]3 years ago

4 0

Answer:

-11a^2+ab+4ab^2

Step-by-step explanation:

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3 years ago

(1 point) Let y=ex/5. Find the differential dy when x=2 and dx=0.2 Find the differential dy when x=2 and dx=0.01

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2 years ago

Evaluate, given f(x) = 6x – 9.<br><br> a) f(4) <br><br> B) f(1/2)

lana66690 [7]

A)f(4)=6(4)-9=24-9=15
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3 years ago

A coach is assessing the correlation between the number of hours spent practicing and the average number of points scored in a g

cricket20 [7]

Answer:

a) $r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1$

We have a perfect linear relationship between the two variables

b) $m=\frac{90}{15}=6$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2$

$\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17$

And we can find the intercept using this:

$b=\bar y -m \bar x=17-(6*2)=5$

So the line would be given by:

$y=6 x +5$

c) For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

Step-by-step explanation:

We have the following data:

Number of hours spent practicing (x) 0 0.5 1 1.5 2 2.5 3 3.5 4

Score in the game (y) 5 8 11 14 17 20 23 26 29

Part a

The correlation coefficient is given:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=9 $\sum x = 18, \sum y = 153, \sum xy = 396, \sum x^2 =51, \sum y^2 =3141$

$r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1$

We have a perfect linear relationship between the two variables

Part b

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=51-\frac{18^2}{9}=15$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=396-\frac{18*153}{9}=90$

And the slope would be:

$m=\frac{90}{15}=6$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2$

$\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17$

And we can find the intercept using this:

$b=\bar y -m \bar x=17-(6*2)=5$

So the line would be given by:

$y=6 x +5$

Part c

For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

5 0

3 years ago