Answer:
Emilio will have both activities again on the same day after 90 days from this Saturday.
Step-by-step explanation:
<em>To solve this question list the multiples of 30 and 9 to find the first common multiple because the two activities will happen again in </em><em>a number divisible by both 30 and 9.</em>
∵ The multiples of 30 are: 30, 60, 90, 120, ............
∵ The multiples of 9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, .....
→ The common multiple is 90
∴ The common multiple of 30 and 9 is 90
→ That means the activities will meet again after 90 days from
this Saturday
Emilio will have both activities again on the same day after 90 days from this Saturday.
Answer:
B. m ∠ 1 = 90° and m ∠ 2 = 90°
Step-by-step explanation:
For most situations, the conjecture would probably be true, but there is one exception that makes this statement false.
When two right angles are supplementary, none of them is acute.
For an angle to be acute it needs to be lesser than 90°, and for a pair of angles to be supplementary they should add up to exactly 180°.
With a pair of right angles (90° each), their sum adds up to 180° but neither of them are acute.
Therefore, the answer is B. m ∠ 1 = 90° and m ∠ 2 = 90°
Answer:
A
Step-by-step explanation:
We can see that Function A's y coordinate doubles every time. The function A = f(x) = 5(2)^x. It is an exponential growth function, and therefore y can never be 0. This means that A does not have an x-intercept.
Function B is a rational function. x cannot be 0, since that would result in an undefined number. This also means that B does not have an x-intercept.
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the following Sum to Product Identities:
<u>Proof LHS → RHS</u>
![\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cqquad%20%5Cdfrac%7B%5Ccos%2010%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7B15%2B25%7D%7B2%7D%5Cbigg%29%5Csin%20%5Cbigg%28%5Cdfrac%7B15-25%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%7D%7B%5Ccos%2020%5Cbigg%5B-2%5Csin%20%5Cbigg%28%5Cdfrac%7B15%2B5%7D%7B2%7D%5Cbigg%29%5Csin%20%5Cbigg%28%5Cdfrac%7B15-5%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%7D)
![\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10](https://tex.z-dn.net/?f=%5Ctext%7BSimplify%3A%7D%5Cqquad%20%5Cqquad%20%5Cdfrac%7B%5Ccos%2010%5B2%5Ccos%2020%5Csin%20%28-5%29%5D%7D%7B%5Ccos%2020%5B-2%5Csin%2010%5Csin%205%5D%7D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D%5Cdfrac%7B-2%5Ccos10%20%5Ccos%2020%20%5Csin%205%7D%7B-2%5Csin%2010%20%5Ccos%2020%20%5Csin%205%7D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D%5Cdfrac%7B%5Ccos%2010%7D%7B%5Csin%2010%7D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D%5Ccot%2010)
LHS = RHS: cot 10 = cot 10 
