Answer:
s= 15 (nearest integer), 14.8 (to nearest 10ths)
Step-by-step explanation:
here you need to apply the cosine rule,
substituting the values,
u= 13, t=5 and angle S as 60 and solving will give you the answer 14.76 (2 dp)
If De'Angelo uses 6 cups of flour and 4 cups of sugar in his birthday cake, how much flour would
2 answers:
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Answer:
The mid-point (p + q , q + p) of AB is the same distance from the x-axis and the y-axis
Step-by-step explanation:
*<em> Lets explain how to solve the problem</em>
- Any point will be equidistant from the x-axis and the y-axis must have
equal coordinates
- Ex: point (4 , 4) is the same distance from the x-axis and the y-axis
because the distance from the x-axis to the point is 4 (y-coordinate)
and the distance from the y-axis and the point is 4 (x-coordinate)
- If (x , y) is the mid-point of a segment its endpoints are
and
, then
and
* <em>Lets solve the problem</em>
∵ Point A has coordinates (p , q)
∵ Point B has coordinates (p + 2q , q + 2p)
- The mid-point of AB is (x , y)
∵
∴
- Take 2 as a common factor from the terms of the numerator
∴
- Divide up and down by 2
∴ x = p + q
∵
∴
- Take 2 as a common factor from the terms of the numerator
∴
- Divide up and down by 2
∴ y = q + p
∴ <em>The mid point of AB is (p + q , q + p)</em>
- p + q is the same with q + p
∵ The x-coordinate of the mid point of AB is p + q
∵ The y-coordinate of the mid point of AB is q + p
∵ p + q = q + p
∴ The coordinates of the mid-point of AB are equal
- According the explanation above
∴ The mid-point (p + q , q + p) of AB is the same distance from the
x-axis and the y-axis
Answer:
5 times as large
Step-by-step explanation:
You can think of "10^5" as "green marbles" if you like. Then your question is ...
5 green marbles is how many times as large as 1 green marble.
Hopefully, the answer is all too clear: it is 5 times as large.
_____
In math terms, when you want to know how many times as large y is as x, the answer is found by dividing y by x:
y/x . . . . . tells you how many times as large as x is y.
Here, that looks like ...