Answer:
Approximately (2, 1)
See image
Step-by-step explanation:
Rearrange each equation so it is ready to be graphed on an x-y-axis (see image) look for the point where the two lines cross. This is the solution for the system of equations.
Answer:
Step-by-step explanation:
Reduction to normal from using lambda-reduction:
The given lambda - calculus terms is, (λf. λx. f (f x)) (λy. Y * 3) 2
For the term, (λy. Y * 3) 2, we can substitute the value to the function.
Therefore, applying beta- reduction on "(λy. Y * 3) 2" will return 2*3= 6
So the term becomes,(λf. λx. f (f x)) 6
The first term, (λf. λx. f (f x)) takes a function and an argument, and substitute the argument in the function.
Here it is given that it is possible to substitute the resulting multiplication in the result.
Therefore by applying next level beta - reduction, the term becomes f(f(f(6)) (f x)) which is in normal form.
Answer:
m(∠C) = 18°
Step-by-step explanation:
From the picture attached,
m(arc BD) = 20°
m(arc DE) = 104°
Measure of the angle between secant and the tangent drawn from a point outside the circle is half the difference of the measures of intercepted arcs.
m(∠C) = ![\frac{1}{2}[\text{arc(EA)}-\text{arc(BD)}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%5Ctext%7Barc%28EA%29%7D-%5Ctext%7Barc%28BD%29%7D%5D)
Since, AB is a diameter,
m(arc BD) + m(arc DE) + m(arc EA) = 180°
20° + 104° + m(arc EA) = 180°
124° + m(arc EA) = 180°
m(arc EA) = 56°
Therefore, m(∠C) = 
m(∠C) = 18°
Answer:
27.20seconds
Step-by-step explanation:
hope this helped,brainliest?