Answer:
(- 9, - 9)
Step-by-step explanation:
Given the 2 equations
3x - 4y = 9 → (1)
- 3x + 2y = 9 → (2)
Adding (1) and (2) will eliminate the term in x
(1) + (2) term by term
(3x - 3x) + (- 4y + 2y) = (9 + 9)
0 - 2y = 18
- 2y = 18 ( divide both sides by - 2 )
y = - 9
Substitute y = - 9 into (1) or (2) and evaluate for x
(1) → 3x + 36 = 9 ( subtract 36 from both sides )
3x = - 27 ( divide both sides by 3 )
x = - 9
Solution is (- 9, - 9)
Divide both sides by 5^2 (which is 25). So p = (5^12)/25. I can't be bothered to plug in into a calculator but be my guest.
is the slope between these two points. To figure out the slope between two points, we need to find the rise over run, which is the change in the <em>y value over the change in the x value</em>. This becomes:
And simplifies to:
, meaning that the slope between these two points is
.
Answer:
11). 
= 90°
12). 
= 212°
Step-by-step explanation:
A circle F with AB and CD are the diameters has been given in the figure attached.
11). Since, 
= 180°
and 
Therefore, 
= 180°
32° + + 58° = 180°
= 180° - 90°
= 90°
12. Since, 
= 32°
= 
= 180° + 32°
= 212°
Answer:
35
Step-by-step explanation:
We know the factors of Lena's age are 2 and 5. The least common multiple must have these factors and the factors of 14, so will at least have factors of 2, 5, and 7.
Apparently, the dad's age is 5·7 = 35.
___
The GCF is 5; the LCM is 70 = 5×14.
_____
Sometimes, I use a little 3-part diagram to think about LCM and GCF. Here, it would look like ...
(2 [5) 7]
where the numbers in curved brackets (2·5) and the numbers in square brackets [5·7] are factors of the two numbers of concern (Lena's age, her dad's age). The middle number in both brackets [5) is the greatest common factor, and the product of all three numbers is their least common multiple.
Here, the product of outside numbers, 2·7 = 14, represents the ratio of the LCM to the GCF. We know that Lena's age has factors of only 2 and 5, so the numbers in the diagram have to be (2[5)7], where 2 and 7 are on the ends and 5 is in the middle.
