Option C:
x = 90°
Solution:
Given equation:
<u>To find the degree:</u>
Subtract 1 + cos²x from both sides.
Using the trigonometric identity:
Let sin x = u
Factor the quadratic equation.
u + 2 = 0, u – 1 = 0
u = –2, u = 1
That is sin x = –2, sin x = 1
sin x can't be smaller than –1 for real solutions. So ignore sin x = –2.
sin x = 1
The value of sin is 1 for 90°.
x = 90°.
Option C is the correct answer.
Not sure I'm right but:
If the bar weighs 9.25 ounces and 65% of the bar is gold, you'd do 65% of 9.25. So the answer would be <span>65% of 9.25= 6.0125</span>
Answer:
37 & 67
Step-by-step explanation:
11 & 33
42 & 56
- 42 - 1, 2, 3, 6, 7, 14, 21, 42
- 56 - 1, 2, 4, 7, 8, 14, 28, 56
- Both 42 and 56 are composite numbers
- 42 & 56 is incorrect
37 & 67
- 37 - 1, 37
- 67 - 1, 67
- Both 37 and 67 are prime numbers
- 37 & 56 is correct
57 & 97
- 57 - 1, 3, 19, 57
- 97 - 1, 97
- 57 is a composite number, 97 is a prime number
- 57 & 97 is incorrect
Answer:
slope is 2/3
Step-by-step explanation:
y=mx + b where m is the slope
Answer: Choice A
The triangles are congruent because both the corresponding sides of the triangles and the corresponding angles of the triangles are congruent.
==============================================
Explanation:
Congruent triangles are identical copies of one another. This means the corresponding pieces must be the same.
It's like saying two houses are identical, so that means the all the various parts (eg: front door, windows, etc) must be identical. If let's say the two front doors were different, then the houses wouldn't be completely identical.
Going back to the triangles, we know that the sides are congruent by the tickmarks
- side MN = side RS (single tickmark)
- side NP = side ST (double tickmarks)
- side MP = side RT (triple tickmarks)
That takes care of the first part of choice A.
And similarly, the angle markers tell us which angles are congruent
- angle M = angle R (single arc)
- angle N = angle S (90 degree angle marker)
- angle P = angle T (double arc)
That concludes the second part of choice A.
