-2x(x-3) = -11
-2x(x) -2x(-3) = -11
-2x² + 6x = - 11
-2x² + 6x + 11 = 0
Let us use the quadratic formula in solving for the value of x.
-2x² + 6x + 11 = 0
a = -2 ; b = 6 ; c = 11
x = (-b <u>+</u> √b² -4ac) ÷ 2a
x = (-6 + √6² - 4(-2)(11)) ÷ 2(-2)
x = (-6 + √36 + 88) ÷ -4
x = (-6 + √124) ÷ -4
x = (-6 + 11.14) ÷ -4
x = 5.14 ÷ -4
x = -1.285 or -1.29
x = (-6 - 11.14) ÷ -4
x = -17.14 ÷ -4
x = 4.285 or 4.29
Answer: 2A/base = Height
Step-by-step explanation:
Step-by-step explanation:
do you need graph paper?
just minus the 2nd cordnates x to the 1dt cordnates x and same for the y. that will give you the distance for each x and y points
T us assume the two numbers to be "x" and "y".
Then
2x + y = 310
And
x - y = 55
Let us take the second equation and find the value of x in relation to y.
x - y = 55
x = y + 55
Now let us put the value of x in the first equation, we get
2x + y = 310
2(y + 55) + y = 310
2y + 110 + y = 310
3y = 310 - 110
3y = 200
y = 200/3
= 66 2/3
Now putting the value of y in the second equation, we get
x - y = 55
x - (200/3) = 55
3x - 200 = 55 * 3
3x = 165 + 200
x = 365/3
= 121 2/3
So the value of x is 121 2/3 and the value of y is 66 2/3
Okay so let’s say the permit fish is the variable p.
The tarpon is 11p.
So, to set up the equation,
(11p)+p=168
Start by combining like terms on the left
12p=168
Divide both sides by 12 to isolate p
p=14
So therefore, the permit fish was 14 pounds.
To find the weight of the tarpon, insert the 14 back into (11p) 11(14)=154.
You can check by adding these together 14+154=168
So:
The permit fish weighed 14 pounds.
The tarpon weighed 154 pounds.
Hope this helps!