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Ask QuestionPosted by Shivam Dey 6 years, 6 months ago
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Posted by Udbhav Raghuvanshi 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We need to make the whole numbers as a rational number by multiplying and dividing the numerator and the denominator with the same number
Thus, We have,
-1 = {tex}-\frac{1}{1}=\frac{-1 \times 10}{1 \times 10}=-\frac{10}{10}{/tex}
1 = {tex}\frac{1}{1}=\frac{1 \times 10}{1 \times 10}=\frac{10}{10}{/tex}
So, {tex}-\frac{9}{10}<-\frac{8}{10}<-\frac{7}{10}<-\frac{6}{10}<-\frac{5}{10}<-\frac{4}{10}<-\frac{3}{10}{/tex}
or - 1 {tex}-\frac{9}{10}<-\frac{8}{10}<-\frac{7}{10}<-\frac{6}{10}<-\frac{5}{10}<-\frac{4}{10}<-\frac{3}{10} < 1{/tex}
{tex}\therefore{/tex} four rational numbers between –1 and 1 are
{tex}-\frac{9}{10},-\frac{8}{10},-\frac{7}{10},-\frac{6}{10}{/tex}
Posted by Sanjaykumar Sanjay Kumar 6 years, 6 months ago
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Posted by Annesha Mondal 6 years, 6 months ago
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Yogita Ingle 6 years, 6 months ago
In ∆ABC, AB = 15 cm , BC = 13 cm and AC = 14 cm.
By Heron's formula ,
S = ( A + B + C ) / 2
S = ( 15 + 13 + 14 ) / 2
S = 42 / 2
S = 21
Now, area of triangle
= √S (S–A) (S–B) (S–C)
= √21 (21–15) (21–13) (21–14)
= √21 (6) (8) (7)
= √7056
= 84 cm2
Again,area of triangle = 1/2×base×altitude=84 cm2
1/2×14×altitude = 84
7 × altitude=84
altitude = 84/7
Altitude = 12cm
Therefore, the area of ∆ABC is 84cm² and its altitude on AC is 12 cm.
Posted by Aalok Rock 6 years, 6 months ago
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Kanchi Singh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
According to the question x is extra milk and y is the expenditure for the month
If the quantity of extra milk be 'x' and expenditure be Rs. 'y' then the given condition.
y = 20x + 500 (As Rs.500 is the fixed expenditure) ...(i)
Put x = 0 in equation (i)
y = 20(0) + 500 = 0 + 500
y = Rs.500
When no extra milk is taken the expenditure remains Rs.500 same
Put y = 1000 in equation (i)
1000 = 20x + 500
1000 - 500 = 20x
500 = 20x
x = {tex}\frac{500}{20}=25 \mathrm{kg}{/tex}
When the 25 kg of milk is taken extra the expenditure increased as Rs.1000
Put x = 2 in equation (i)
y = 20(2) + 500
y = 40 + 500
y = Rs. 540
When the 2 kg of milk is taken extra the expenditure increased by Rs.40 i.e. Rs. 540
| 0 | 25 | 2 |
| 500 | 1000 | 540 |
Posted by Malkeet Singh 6 years, 6 months ago
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Posted by Manish Chawan 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
- The mirror image of point (2, 3) is (2, -3) with respect to x-axis.
- The mirror image of (-4, -6) is (-4,6) with respect to x-axis.
Priya Bhagoria 6 years, 6 months ago
Yogita Ingle 6 years, 6 months ago
In image of a point P(x, y) with respect to x axis, the change in only sign of ordinate of point so the image is Q (x, -y)
So, the mirror image of the point (2,3) is (2 ,-3) and (-4,-6) is (-4,6)
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Yogita Ingle 6 years, 6 months ago
The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations.
Median: Let’s arrange their score in ascending order in blue table. Now we will take the middle most score. Since we are talking about central tendency of data, middle most score should also reflect central tendency of data.
When the number of observations (n) is odd, the median is the value of the ((n+1)/2) Th observation.
When the number of observations (n) is even, the median is the mean of (n/2) & (n/2 + 1) Th observations.
Posted by Pankaj Tyagi 6 years, 6 months ago
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Gaurav Seth 6 years, 6 months ago
In point A(1, -1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant. In point B(4, 5), both coordinates are positive, so it lies in I quadrant. On plotting these point, we get the following graph.
(i) On joining the points A and B, we get the line segment AB. Now, to find the coordinates of a point on this line segment between A and B draw a perpendicular to X-axis from x = 2 and 3.
[since, x = 2 and 3 lies between A and B] say it intersect line segment AB at P and p’. Now, draw a perpendicular to Y-axis from P and p’, they intersect Y-axis at y = 1 and 3, respectively. Thus, we get points (2,1) and (3, 3) which lie between line segment AB.
(ii) Extent the line segment AB. Now, draw a perpendicular to X-axis from x = 5, say it intersects extended line segment at Q. Again, draw a perpendicular to Y-axis from Q and it intersects Y-axis at y = 7. Thus, we get the point Q(5,7) which lies outside the line segment AB.
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Bhoomika . 6 years, 6 months ago
3Thank You