#3 is < , =, =
#6 4L= 4,000ml
9L= 9,000ml
6g=6000mg
Answer:
17
Step-by-step explanation:
it would technically be 16.5 because if u do 1,900/115=16.521739043 but just round it up to 17 so there you go I hope this helps you
If we want to find when the population of species A will be equal to the population of species B, we need to see when the two equations for the population of each species are equal, ie. equate them and solve for t. Thus:
2000e^(0.05t) = 5000e^(0.02t)
(2/5)e^(0.05t) = e^(0.02t) (Divide each side by 5000)
2/5 = e^(0.02t) / e^(0.05t) (Divide each side by e^(0.05t))
2/5 = e^(-0.03t) (use: e^a / e^b = e^(a - b))
ln(2/5) = -0.03t (use: if b = a^c, then loga(b) = c )
t = ln(2/5) / -0.03 (Divide each side by -0.03)
= 30.54 (to two decimal places)
Therefor, the population of species A will be equal to the population of species B after 30.54 years.
I wasn't entirely sure about the rounding requirements so I've left it rounded to two decimal places.
Answer:
X=-14.5 Y=3.2
Step-by-step explanation:
5x-13=5y+7x
subtract 7x from both sides
-2x-13=5y
subtract 5y from both sides
-2x-5y-13=0
add 13 to both sides
-2x-5y=13
<u>SYSTEMS OF EQUATIONS NOW</u>
4x+15y=-10
-2x-5y=13
multiply the second equation by 3
4x+15y=-10
-6x-15y=39
add the equations together to get:
-2x=29
divide both by -2
x=-14.5
Substitute x into one equation
4(-14.5)+15y=-10
-58+15y=-10
add 58 to both sides
15y=48
divide both sides by 15
y=3.2
<em>CHECK YOUR WORK</em>
4(-14.5)+15(3.2)=-10
-58+48=-10
-10=-10
If Mr. Sanchez has 16 feet of fencing, the perimeter of his fencing must be 16 feet too. The perimeter is the sum of all the side lengths; since he has 16 feet of fencing, this is all he can use.
He wants the fencing to be a rectangle. The area of a rectangle is equal to length times width: A=lw. The perimeter of a rectangle is equal to the length plus width plus length plus width, or twice the length plus twice the width: P=2l+2w.
We know that the perimeter (P) must equal 16. We can fill this in: 16=2l+2w. What are some possible lengths and widths that would fulfill this? What if l=1? Then we can fill that in: 16=(2*1)+2w. 2*1=2. So, 16=2+2w. Subtract 2 from both sides to get 14=2w. Then divide both sides by 2: w=7. This is a possible fencing scenario: the length is 1, and the width is 7. BUT, we need to get the greatest area possible. The area of this rectangle is A=lw, which would be A=1*7. The area of this fenced area would be 7 feet squared.
What if l=2? Use a similar method. 16=(2*2)+2w. 2*2=4. 16=4+2w. 12=2w. w=6. A=lw. A=2*6=12 ft squared.
l=3: 16=(2*3)+2w. 2*3=10. 16=6+2w. 10=2w. w=5. A=3*5=15 sq. ft.
l=4: 16=(2*4)+2w. 2*4=8. 16=8+2w. 8=2w. w=4. A=4*4=16 sq ft.
After this, it starts to repeat: if l=5, then w=3, and you get the same area as l=3: 15. If l=6, w=2, and A=12. If l=7, w=1, and A=7. l=8 or w=8 is impossible, because then the other dimension is 0 and you get an area of 0. You can't have any dimensions above 8, either, or you will get a perimeter greater than 16--and then your other dimension would be negative, which is impossible in length.
So the possible areas are 7, 12, 15, and 16. What's the greatest area? 16.
