Answer:

Step-by-step explanation:
Let w and l be the width and length of the garage.
Let
and
be the area of garage at present and new garage.
Given:
The area of the garage at present

And he planed to tripling the dimensions of the garage.
We need to find the area of the new garage.
Solution:
We know the area of the rectangular garage.


---------------(1)
The dimension of the new garage is triple, so the area of the new garage is.


Substitute
from equation 1.


Therefore, the area of the new garage 
Answer:
The solutions are


Step-by-step explanation:
we have

Group terms that contain the same variable, and move the constant to the opposite side of the equation

Factor the leading coefficient

Complete the square. Remember to balance the equation by adding the same constants to each side


Rewrite as perfect squares


square root both sides




Multiply the first equation by 7 and the second equation by 2, and then add.
Step-by-step explanation:
2f - 5g=-9
-7f + 3g=4
To eliminate any variable , we need to make the coefficients same with different sign and add it
the coefficient of 'f' in first equation is 2
the coefficient of 'f' in second equation is -7
We already have different sign so we make the coefficients same
To make the coefficient same , we multiply the first equation by 7 and second equation by 2
So equation becomes
14f - 35g = -63
-14f + 6g = 8
Now when we add both equations, 'f' gets cancelled
-29g = -55
Answer:
a) D(t) = ![\[20*\sqrt{5} * t\]](https://tex.z-dn.net/?f=%5C%5B20%2A%5Csqrt%7B5%7D%20%2A%20t%5C%5D)
b) 178.885 miles
Step-by-step explanation:
Ship A travels north at the rate of 20 mph.
Ship B travels east at the rate of 40 mph.
After t hours, Ship A is at a distance of 20t miles from the origin.
Similarly, Ship B is at a distance of 40t miles from the origin.
(a) Distance D(t) = ![\[\sqrt{(20t)^{2}+(40t)^{2}}\]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B%2820t%29%5E%7B2%7D%2B%2840t%29%5E%7B2%7D%7D%5C%5D)
= ![\[\sqrt{400*t^{2}+1600 * (t)^{2}}\]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B400%2At%5E%7B2%7D%2B1600%20%2A%20%28t%29%5E%7B2%7D%7D%5C%5D)
= ![\[\sqrt{2000*t^{2}}\]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B2000%2At%5E%7B2%7D%7D%5C%5D)
= ![\[20*\sqrt{5} * t\]](https://tex.z-dn.net/?f=%5C%5B20%2A%5Csqrt%7B5%7D%20%2A%20t%5C%5D)
(b) Distance between the two ships when t = 4,
= ![\[20*\sqrt{5} * 4\]](https://tex.z-dn.net/?f=%5C%5B20%2A%5Csqrt%7B5%7D%20%2A%204%5C%5D)
=
miles
= 80 * 2.236
= 178.885 miles
<span>Using PEMDAS, solve the equation
−2\3 (3- 1\2) (-1) -> -2/ 3(3-0.5)(-1) -> -2/3(2.5)(-1) - > -2/3(-2.5) -> 5/3 -> 1 2/3
1 and 2/3 is the same thing as 2-1/3
The answer is B
</span>