Answer:
0
Step-by-step explanation:
any log with a base of one and it becomes logv5 (1) after logv5(logV3 (3) because log3(3) equal one so then logv5 (1) is 0
<u><em>Answer:</em></u>
SAS
<u><em>Explanation:</em></u>
<u>Before solving the problem, let's define each of the given theorems:</u>
<u>1- SSS (side-side-side):</u> This theorem is valid when the three sides of the first triangle are congruent to the corresponding three sides in the second triangle
<u>2- SAS (side-angle-side):</u> This theorem is valid when two sides and the included angle between them in the first triangle are congruent to the corresponding two sides and the included angle between them in the second triangle
<u>3- ASA (angle-side-angle):</u> This theorem is valid when two angles and the included side between them in the first triangle are congruent to the corresponding two angles and the included side between them in the second triangle
<u>4- AAS (angle-angle-side):</u> This theorem is valid when two angles and a side that is not included between them in the first triangle are congruent to the corresponding two angles and a side that is not included between them in the second triangle
<u>Now, let's check the given triangles:</u>
We can note that the two sides and the included angle between them in the first triangle are congruent to the corresponding two sides and the included angle between them in the second triangle
This means that the two triangles are congruent by <u>SAS</u> theorem
Hope this helps :)
Answer:
She's been charged 84 dollars
Step-by-step explanation:
Let's call each month m
if every month it deducts 7 dollars, then to find how much has been charged after a year, we do
-7 * 12 = -84
Answer:
8
Step-by-step explanation:
The second one because for every x value there is one and only one y value. If you plotted the points and graphed it, you would know it is not a function if it doesn't pass the vertical line test. Notice the same x values show up repeatedly in the other ordered pairs with different y values. Only one y value for every x value
