According to the figure, I can safely assume that the 10 cm line and 15 cm line are parallel.
Thus, there are two similar triangles, one with the 10 cm bottom and 15 cm bottom, where one triangle is larger by a factor of 15/10 = 3/2
We know that the 8 cm line segment and y both join to create the side of the large triangle.
8cm + y = the side of the large triangle
Multiplying 8cm by 3/2 gives us 12, since we know the large triangle is 3/2 times larger than the smaller one.
8 + y = 12
y = 4 cm
Finding x is going to be a bit different. We know that the 6 cm line and x form the side of the larger triangle, which we know is 3/2 times larger than x.
x + 6cm = ?
The side of the larger triangle is 3/2x, thus
x + 6 = 3/2 x
Subtract both sides by x
6 = 1/2 x
Multiply both sides by 2
12 cm = x
Thus, y = 4 and x = 12.
Let me know if you need any clarifications, thanks!
Answer: None the zeroes have a multiplicity of 2.
Step-by-step explanation: i took the test
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
Answer:
ABCDEFGHIJKLMNOPQRSTUFWTYZ
Step-by-step explanation:
F(x) = 3x - 16x - x + 4
f(x) = - 14x + 4
I can not see your picture
