<em>Greetings from Brasil...</em>
According to the statement of the question, we can assemble the following system of equation:
X · Y = - 2 i
X + Y = 7 ii
isolating X from i and replacing in ii:
X · Y = - 2
X = - 2/Y
X + Y = 7
(- 2/Y) + Y = 7 <em>multiplying everything by Y</em>
(- 2Y/Y) + Y·Y = 7·Y
- 2 + Y² = 7X <em> rearranging everything</em>
Y² - 7X - 2 = 0 <em>2nd degree equation</em>
Δ = b² - 4·a·c
Δ = (- 7)² - 4·1·(- 2)
Δ = 49 + 8
Δ = 57
X = (- b ± √Δ)/2a
X' = (- (- 7) ± √57)/2·1
X' = (7 + √57)/2
X' = (7 - √57)/2
So, the numbers are:
<h2>
(7 + √57)/2</h2>
and
<h2>
(7 - √57)/2</h2>
<h3>Answer:</h3><h3>41</h3><h3>Step-by-step explanation:</h3><h3>multiplication and addition</h3><h3>5x4=20</h3><h3>7x3=21</h3><h3>20+21=41</h3>
5x-2+x-3+x-3 = 34
7x-8 = 34
7x/7 = 42/7
x = 6
Answer:
The correct answer is 24
Step-by-step explanation:
to solve this you will need to use the pathagreom theorum
a^{2}+b^{2}=c^{2}
A= one side lenth
B= the secons side lenth
C= hypotnuse
It is helpfull to draw out the situation
you know that the latter is 25 ft, that is your hypotnuse
you also know that the 7 ft away from the base of the building is one of the side lenths, lets call it side a
so plug the numbers into the equation
7^2 + b^2 = 25 ^2
you leave b^2 alone because that is the side you are trying to find
now square 7 and 25 but leave b^2 alone
49 + b^2 = 625
now subtract 49 from both sides
b^2 = 576
now to get rid of the square of b you have to do the opposite and square root both sides removing the square of the B and giving you the answer of..........
B= 24
Hope this helped!! I tryed to explain it as simpil as possiable
Answer:
The answer is 4
Step-by-step explanation:
1/2+(-5/6)= 1 1/3
1 1/3 + 1/4= 1 7/12
