You would use PEMDAS
But since any number divided by 1 is the number by which it is divided by then you can eliminate that. Now your problem looks like this:
8✖️4➕9✖️56➕57
Since there are no parentheses or exponents then you would multiply.
8✖️4= 32
32➕9✖️56➕57
Then multiply the other multiplication problem.
9✖️56= 504
32➕504➕57
Now you can add all of the numbers together
32➕504=536
536➕57=593
Your answer is 593
Hope this helps! :3
Answer:
3√(22) + 7
Step-by-step explanation:
The way this is worded can be interpreted two different ways so just to cover the bases, I'll do both. I'm confident they are asking for way 1 though because it wants you to simplify the expression, not evaluate. Way 2 gives you a solution rather than an exact equation.
<u>Number 1</u>
√(6) x √(33) + 7
= √(6 x 33) + 7
= √(198) + 7
= √(9 x 22) + 7
= 3√(22) + 7
<u>Number 2</u>
√(6) x √(33 + 7)
= √(6) x √(40)
= √(6) x (4√(10))
= 4√(60)
= 8√(15)
Answer:
B
Step-by-step explanation:
use quadratic formula: (-b ±√b²-4ac) ÷ 2a
a = 10, b = -9, c = -6
The following formula is used to find the answer.
D = 50 mg (0.6^n)
D is the dosage
n is at any hour
Using this formula and solving the equation for it, the answer is 18.
Answer:
Step-by-step explanation:
The hyperbola has x-intercepts, so it has a horizontal transverse axis.
The standard form of the equation of a hyperbola with a horizontal transverse axis is 
The center is at (h,k).
The distance between the vertices is 2a.
The equations of the asymptotes are
1. Calculate h and k. The hyperbola is symmetric about the origin, so
h = 0 and k = 0
2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6
a = 6/2 = 3
3. For 'b': The equation for the asymptote with the positive slope is
Thus, asymptote has the slope of
4. The equation of the hyperbola is
The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.
