Answer:
1 ± √47
Step-by-step explanation:
Combine like terms in 2x^2 +3 x-7 =x^2 +5x +39:
2x^2 - x^2 = x^2 (first term);
3x - 5x = -2x (second term);
-7 - 39 = -46 (third term)
Then we have, all on the left side, x^2 - 2x - 46, which is a quadratic equation. Here the coefficients are a = 1, b = -2 and c = -46.
Then the discriminant, b^2 - 4ac, is:
(-2)^2 - 4(1)(-46) = 4 + 184.
The roots are:
-(-2) ±√188 2 ± √4√47
------------------- = -------------------
2 2
= 1 ± √47 (last of the answer choices)
=
x = ------------------
When expanding number, we should separate the numbers according to the places like the number 275 written in expanded form 200+70+5. So, in this question we must likely done like this. The number is 39,005 the sum that is written on expanded form is 30,000+9,000+5.
Answer:
A number line going from negative 7 to negative 1. A closed circle is at negative 5. Everything to the right of the circle is shaded.
Step-by-step explanation:
-4.4 ≥ 1.6 x - 3.6
Solving for the value of x,
1.6 x ≤ -4.4 + 3.6
1.6 x ≤ - 0.8
Dividing 1.6 from both sides to know value of x
x ≤ - 0.5
Answer:
39.5 ft
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the relation between angles and sides of a right triangle.
Tan = Opposite/Adjacent
This lets us write two equations in two unknowns:
tan(67°) = AD/CD . . . . . . . . . . angle at guy point
tan(39°) = AD/(CD+32) . . . . . .angle 32' farther
__
Solving the first equation for CD and using that in the second equation, we can get an equation for AD, the height of the tower.
CD = AD/tan(67°)
tan(39°)(CD +32) = AD . . . . eliminate fractions in the second equation
tan(39°)(AD/tan(67°) +32) = AD
32·tan(39°) = AD(1 -tan(39°)/tan(67°)) . . . simplify, subtract left-side AD term
32·tan(39°)tan(67°)/(tan(67°) -tan(39°)) = AD . . . . divide by AD coefficient
AD ≈ 39.486 . . . . feet
The tower is about 39.5 feet high.
Answer:
0.033 ft
Step-by-step explanation:
Let g = 32.2 ft/s2 and h be the maximum height that water can be filled before the sides shatter.
Pressure is distributed from 0 to maximum at the bottom like the following equation:

So the force generated by water pressure on the side of the tank is

where s = 6ft is the side length of the tank. This force cannot be larger than 200lb


