Answer:
m∠RQS = 72°
m∠TQS = 83°
Step-by-step explanation:
m∠RQS +m ∠TQS = m∠RQT
The two angles combine to make a larger angle
So
m∠RQS = (4x - 20)
m∠TQS = (3x + 14)
(4x - 20) + (3x + 14) = 155
Group the Xs and the constants
4x + 3x - 20 + 14 = 155
Combine like terms
7x - 6 = 155
Add 6 to both sides
7x = 161
Divide by 7 on both sides
x = 23
Check:
4(23) - 20 + 3(23) + 14 = 155
92 - 20 + 69 + 14 = 155
155 = 155
But we need to find m∠RQS and m∠TQS. So plug in x = 23 to the values.
m∠RQS = 4(23) - 20 = 72°
m∠TQS = 3(23) + 14 = 83°
Checking:
72 + 83 = 155
Answer:
Option C 7 is your answer ☺️☺️☺️
When roots of polynomials occur in radical form, they occur as two conjugates.
That is,
The conjugate of (a + √b) is (a - √b) and vice versa.
To show that the given conjugates come from a polynomial, we should create the polynomial from the given factors.
The first factor is x - (a + √b).
The second factor is x - (a - √b).
The polynomial is
f(x) = [x - (a + √b)]*[x - (a - √b)]
= x² - x(a - √b) - x(a + √b) + (a + √b)(a - √b)
= x² - 2ax + x√b - x√b + a² - b
= x² - 2ax + a² - b
This is a quadratic polynomial, as expected.
If you solve the quadratic equation x² - 2ax + a² - b = 0 with the quadratic formula, it should yield the pair of conjugate radical roots.
x = (1/2) [ 2a +/- √(4a² - 4(a² - b)]
= a +/- (1/2)*√(4b)
= a +/- √b
x = a + √b, or x = a - √b, as expected.
Yeah I'm not sure but I think it's 6
Since 1A claims that the diagram is of a square, you can easily find the perimeter by multiplying just one side by 4, because the definition of a square says that all of its four sides are equal in length.
Take the left side, x and 4, and add them together, because both of these lengths add up to form the side of the square. You have found one side of the square, x + 4. Now multiply this side by 4 for the perimeter.
Perimeter is the length all around the figure, and since a square has 4 sides you would multiply one side by 4 to find the perimeter.
4(x + 4) is your expression for the perimeter of the square. You could probably solve 1B and 1C by substituting in 3 and 5 for x in the equation I've given you :)
