1)Find the third side of a triangle when two sides of the triangle is 4 and the included angle measuring 89 degrees 2)you were asked by your teacher to add up the numbers 1 to 100. Please show all work

3)determine if the foci for the following ellipse are correct and explain your reasoning 7x2 + 8y2 = 30; foci: (sqrt11, 0), (-sqrt11, 0)

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)).

well what you have to do is divide 84 by 6 so theanswer

step-by-step explanation:

84 divided by 6 = 14

and that lightbulb changing party must be freaking lit!

the answer to your question is b.