The answer is (-1,7)
I got this answer because to solve a system, you find the point on the graph where both lines intersect. And that point in this graph is (-1,7). Hope this helps!:)))))
Answer:
They will need 200kg of pine logs.
Gauge pressure in the tank = 24045.29 Pa
Absolute pressure= 125370.29Pa
Step-by-step explanation:
Full question:sp.gr=0.8.The fluid in the nanometer is ethyl iodide with sp.gr =1.93. The manometric fluid height is 50 inches. What is the gauge pressure and absolute pressure in the tank?
Converting 50inches to metres. Multiply by 0.0254
50×0.0254=1.27m
Gauge pressure in tank is given by:
P = pgh
Where p= density,g= acceleration due to gravity, h= height
P= 1.93× 1000× 9.81× 1.27= 24045.29Pa
Absolute pressure=Ptotal= Pliquid + Patm
Absolute pressure =24045.29 + 101325=125370.29pa
Answer:
10
Step-by-step explanation:
Each box is worth 2 units, because the x and y axis are counting by 2s
Answer:
c. 28 liters
Step-by-step explanation:
Given tha tJanet is mixing a 15% glucose solution with a 35% glucose solution. This mixture produces 35 liters of a 19% glucose solution. Now we need to find about how many liters of the 15% solution is Januet using in the mixture.
Let the number of liters of the 15% solution is Januet using in the mixture = x
Let the number of liters of the 35% solution is Januet using in the mixture = y
Then we get equations:
x+y=35...(i)
and
(15% of x) + (35% of y) = 19% of 35.
or
0.15x+0.35y=0.19(35)
15x+35y=19(35)
3x+7y=19(7)
3x+7y=133 ...(ii)
solve (i) for x
x+y=35
x=35-y...(iii)
Plug (iii) into (ii)
3x+7y=133
3(35-y)+7y=133
105-3y+7y=133
105+4y=133
4y=133-105
4y=28
y=28/4
y=7
plug y=7 into (iii)
x=35-y=35-7=28
Hence final answer is c. 28 liters
Answer: The area of the triangle ABC is 168 cm square.
Explanation:
Let the area of the triangle ABC be x cm square.
It is given that In △ABC point D is the midpoint of AB
, point E is the midpoint of BC
, and point F is the midpoint of BE
.
As we know that a median of a triangle divides the area of a triangle in two equal parts. Since D is a midpoint of AB and AD is median of triangle ABC. So area of triangle ACD and BCD is half of the area of triangle ABC.
The area of ACD and BCD is 
.
Since E is a midpoint of BC and DE is median of triangle BCD. So area of triangle BDE and CDE is half of the area of triangle BCD.
The area of BDE and CDE is 
.
Since F is a midpoint of BE and DF is median of triangle BDE. So area of triangle BDF and DEF is half of the area of triangle BDE.
The area of BDF and DEF is 
. As shown in below figure.
It is given that the area of △DCF= 63 cm.
From the figure the area of △DCF is,
Therefore the area of triangle ABC is 168 cm square.
