Write an expression for “seven less than three halves of z”
1 answer:
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Answer:
1) The length of the third side is 5.607 units
2) The sum of the numbers from 1 to 100 is 5050
3) For the x-axis, foci: ((√15/28), 0) and (-(√15/28), 0)
For the y-axis, foci: (0, (√15/28)) and (0, -(√15/28))
Step-by-step explanation:
1) When two sides of the triangle are equal to 4 then the triangle is an isosceles triangle
Given that the included angle (the angle between the two sides) is 89°, we have;
The other two base angles are equal to {180 - 89)/2 = 91/2 = 45.5°
Therefore, we have from cosine rule;
a² = b² + c² - 2·b·c·cos(A)
We note that the angle opposite the third side is the included angle 89°, therefore, when we put a as the third side in the above equation, we have;
a² = 4² + 4² - 2×4×4×cos(89°)
a² = 31.44
a = 5.607
The length of the third side is 5.607 units
2) The numbers 1 to 100 form an arithmetic series with the first term, a = 1 and the common difference, d = 1 with the number of terms n = 100
The sum of an arithmetic progression, Sₙ, is given as follows;
Therefore, by plugging in the values, we have;
Sₙ = 100/2*(2*1 + (100 - 1)*1) = 100/2*(101) = 5050
The sum of the numbers from 1 to 100 is 5050
3) The foci of an ellipse 7·x² + 8·y² = 30 is found as follows;
Dividing both sides of the equation by 30 gives;
7/30·x² + 8/30·y² = 30/30
7/30·x² + 8/30·y² = 30/30
7/30·x² + 4/15·y² = 1
Which is of the form;
x²/a² + y²/b² = 1
For the x-axis we have
c² = a² - b²
c² = 30/7 - 15/4 = 15/28
h = 0, k = 0
Foci: ((√15/28), 0) and (-(√15/28), 0)
For the y-axis, we have;
x²/b² + y²/a² = 1
The foci are then (0, (√15/28)) and (0, -(√15/28)).