Answer:
\frac{13+\left(-3\right)^2+4\left(-3\right)+1-\left(-10-\left(-6\right)\right)}{\frac{\left(4+5\right)}{\left(4^2-3^2\left(4-3\right)-8\right)}+12}=5
Step-by-step explanation:
<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
Answer: 9+4n-1 = 20
We can solve this by substitution.
Replace n with the value given, 3 (remember 4n means 4 times n):
9 + 4*3 - 1
Then work it out using arithmetic
9+12-1
=20
If the greater integer is m, then the smaller integer must be (m-3) because the integers are 3 units apart on a number line. It cannot be in front of m, because m is the greater so it must be behind m meaning it is 3 units less. Product is the technical term for the result of a multiplication problem so m(m-3) = 108 would be the answer.
Final Answer: m(m-3) = 108
Answer:
- ABCD is a rhombus, and a parallelogram
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<h3>Given </h3>
- Points A(-6, - 1), B(4, - 6), C(2, 5), D(- 8, 10)
First, plot the points (see attached picture).
Then, connect all the points.
<h3>We see that:</h3>
- Opposite sides are parallel,
- Diagonals are perpendicular.
From our observation the figure is rhombus.
Let's confirm it with the following.
1) Find midpoints of diagonals and compare.
- AC → x = (- 6 + 2)/2 = - 2, y = (- 1 + 5)/2 = 2
- BD → x = (4 - 8)/2 = - 2, y = (- 6 + 10)/2 = 2
The midpoint of both diagonals is same (- 2, 2).
2) Find slopes of diagonals and check if their product is -1, this will confirm they are perpendicular.
- m(AC) = (5 - (-1))/(2 - (-6)) = 6/8 = 3/4
- m(BD) = (10 - (-6))/(-8 - 4) = - 16/12 = - 4/3
- m(AC) × m(BD) = 3/4 * (- 4/3) = - 1
<u>Confirmed.</u>
So this is a rhombus and also a parallelogram but <u>not</u> rectangle or square, since opposite angles are not right angles.