Answer:
x = 24.52
Step-by-step explanation:
Since this is a right triangle, use Pythagorean Theorem to solve for the hypotenuse.
Pythagorean Theorem: a² + b² = c²
24² + 5² = x²
576 + 25 = x²
601 = x²
√601 = x
The equation has one extraneous solution which is n ≈ 2.38450287.
Given that,
The equation;
We have to find,
How many extraneous solutions does the equation?
According to the question,
An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.
To solve the equation cross multiplication process is applied following all the steps given below.
The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y
with 0 and solve for x. The graph of the equation is attached.
n ≈ 2.38450287
Hence, The equation has one extraneous solution which is n ≈ 2.38450287
For more information refer to the link.
brainly.com/question/15070282
You have 3hours 25seconds left to assemble the machine.
Now cos⁻¹(0.7) is about 45.6°, that's on the first quadrant.
keep in mind that the inverse cosine function has a range of [0, 180°], so any angles it will spit out, will be on either the I quadrant where cosine is positive or the II quadrant, where cosine is negative.
however, 45.6° has a twin, she's at the IV quadrant, where cosine is also positive, and that'd be 360° - 45.6°, or 314.4°.
now, those are the first two, but we have been only working on the [0, 360°] range.... but we can simply go around the circle many times over up to 720° or 72000000000° if we so wish, so let's go just one more time around the circle to find the other fellows.
360° + 45.6° is a full circle and 45.6° more, that will give us the other angle, also in the first quadrant, but after a full cycle, at 405.6°.
then to find her twin on the IV quadrant, we simply keep on going, and that'd be at 360° + 360° - 45.6°, 674.4°.
and you can keep on going around the circle, but only four are needed this time only.
