A. y=2x^3 ----------<span>nonlinear
</span><span>
B. y=x+5 ..........</span><span>linear
</span><span>
C. y=Square root of x-3 </span>----------nonlinear
<span>
D. 5x+5y=25</span>..........linear
<span>
E. y=x^2/2 </span>----------nonlinear
Answer:
D)4x + 6/(x + 1)(x - 1)
Step-by-step explanation:
A field is basically a rectangle, so to find the perimeter of our field we are using the formula for the perimeter of a rectangle
where
is the perimeter
is the length
is the width
We know from our problem that the field has length 2/x + 1 and width 5/x^2 -1, so and .
Replacing values:
Notice that the denominator of the second fraction is a difference of squares, so we can factor it using the formula where is the first term and is the second term. We can infer that and . So, . Replacing that:
We can see that the common denominator of our fractions is . Now we can simplify our fraction using the common denominator:
We can conclude that the perimeter of the field is D)4x + 6/(x + 1)(x - 1).
4x - 7y = 1
if y = 5
4x - (7*5) = 1
4x - 35 = 1
4x = 36
x = 9
other coordinate is 9
Base and height both measure 6cm.
Answer:
x=11, y=1 or (11, 1)
Step-by-step explanation:
I used substitution to solve this. Basically wherever there's an x, I would fill it in as 12-y.
2(12-y)+3y=29
24+2y+3y=29
24+5y=29
subtract 24 from both sides
5y=5
divide by 5
y=1
Now substitute y back in to get x.
x=12-1
x=11
You can also graph this into desmos, and wherever they intercept, that's your answer. I hope this helps!