To solve this, all you have to do is add 1/5 and 1/3.
Well, you might be thinking, "But the denominators are not the same! Fractions have to have like denominators in order to be added together!" And you are right. So, all we have to do, is make them both have the same denominator. To do this, we have to multiply both the numerator and denominator by the same number to find an equivalent fraction.
For 1/5, we can multiply both the numerator and denominator by 3, to get 3/15.
Likewise, we can multiply both the numerator and denominator of 1/3 by 5, to get 5/15.
Now, you can easily add 3/15 and 5/15 because all you have to do is add the numerators, because the denominators are the same!
3/15+5/15=8/15, so she spends 8/15 hour on her math and social studies homework.
Hope I helped!
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
If you meant:
64.3-3*(2³)/2
That would be 52.3 :)
Answer:
(a) 315°
(b) 3°
(c) 238°
Step-by-step explanation:
Bearings are measured clockwise from north. The triangle described is illustrated in the attachment.
<h3>(a)</h3>
The bearing of P from R is 180° different from the bearing of R from P it will be ...
135° +180° = 315° . . . . bearing of P from R
__
<h3>(b)</h3>
The bearing of Q from R is 48° more than the bearing of P from R, so is ...
315° +48° = 363°, or 3° . . . . bearing of Q from R
__
<h3>(c)</h3>
The angle QPR has a value that makes the sum of angles in the triangle equal to 180°. It is ...
180° -48° -55° = 77°
The bearing of Q from P is 77° less than the bearing of R from P, so is ...
135° -77° = 58°
As above, the reverse bearing from Q to P is ...
58° +180° = 238° . . . . bearing of P from Q
Answer:
D
Step-by-step explanation:
A function is where each input (here, the input is x) corresponds to exactly one output (here, the output is y). In other words, if a function is graphed, we should be able to draw a vertical line through every single part of it that will intersect it at only one place.
Let's examine each choice.
(A) Well, if we draw a vertical line through the graph, it will obviously intersect the entire line - which is an infinite number of intersections, so this is not a function.
(B) If we draw a vertical line through the portion of the graph that lies near the positive x-axis, we note that it will intersect twice, so this is not a function.
(C) If we strategically draw a vertical line through the y-axis, we see it will intersect two times, so this is not a function.
(D) We can draw a vertical line through any portion of this graph and know that it will only intersect once.
Therefore, the answer is D.
