Answer:
60
Step-by-step explanation:
Solution:
we are given that
The magnitudes of the moon and Sirius are -12.5 and -1.44 respectively.
we have been asked to find approximately how many times is the moon brighter than Sirius?
The word magnitude is used by the scientist to describe the brightness of moon/star/object etc.
So we can find the brightness by dividing the given values. so we get

Hence we can say
The moon is 8.68 times brighter than Sirius.
The original price was 725.
Explanation:
"36% less than" means taking 36% away from 100%. 100-36 is 64, so 64% remains when 36% is taken away. So, 64% of the original table's price is 464.
So, if the original price was x, 464=0.64x
Solve this by dividing both sides by 0,64:
725=x
So, the original price was 725.
243 liters represents 7/10 - 1/4 of the water.
v=243 / (7/10 - 1/4) = 243 / (14/20-5/20) = 243(20)/9 = 540 liters.
Answer: 540 liters
Check:
(7/10)(540) = 7(54) = 378
(1/4)540 = 135
difference = 378 - 135 = 243 good
Answer:
1. Given
2, Exterior sides on opposite rays
3. Definition of supplementary angles
4. If lines are ||, corresponding angles are equal
5. Substitution
Step-by-step explanation:
For the first one, it is given as shown in the problem. Also in the figure you can see that line s is parallel to line t.
2. ∠5 and ∠7 are adjacent, they share a common side. Their non-common side are rays that go in a direction opposite of each other. Also you can see that they form a straight line, which means that they are supplementary.
3. Supplementary angles simply put are angles that sum up to 180°. You know this for sure because of proof 2, specifically the part that they form a straight line. The measure of a straight line is 180°.
4. Corresponding angles are congruent. These are angles that have the same relative position when a line is intersected by parallel lines. You have other example in the figure like ∠2 and ∠6; ∠3 and ∠7.
5. This is substitution because ∠1 substituted ∠5 in this case. Since ∠1 is equal to ∠5, then it can substitute it in the equation given in step 3. This means that ∠1 and ∠7 are supplementary as well.