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nignag [31]
2 years ago
13

42 and 63 as product of their prime factors what is it?

Mathematics
2 answers:
bekas [8.4K]2 years ago
5 0
42 = 2 x 3 x 7
63 = 3 x 3 x 7
yuradex [85]2 years ago
4 0
Hello,
42=2×3×7 and 63=3×3×7

bye :-)
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Solve the inequality
trapecia [35]

Answer:

x>-1

Step-by-step explanation:

Subtract 0.7 from both sides

6 0
3 years ago
What are the sine and cosine ratios for a 30°–60°–90° triangle with a hypotenuse of 6, when θ = 60°?
Margaret [11]

Answer:

There are two ways to do this problem algebraically or trigonometrically.

Algebraically/geometrically

The ratios of the sides of a 30/60/90 tri. are x, x√3, 2x (short leg, long leg, hyp). Therefore, if the long leg is 6 'units'. then 6 = x√3. x = 6√3.

The hyp is 2x then the hypotenuse is 2(6√3) = 12√3, rationalizing is 12√3/3 = 4√3

Using Trig.

We can use sinx = y/r = opp/hyp. The long leg of 6 is opposite 60 degrees (pi/3).

Therefore, sin(pi/3) = 6/r =

r = 6/sin(pi/3) = 6/(√3/2) = 12/√3, when you rationalize you get 12√3/3 = 4√3

4 0
2 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

7 0
3 years ago
What is 1/5 divided by 1/2
Len [333]
The answer to your question is 0.1
4 0
2 years ago
Read 2 more answers
What is the percent increase 41,700 and 63,00
valentinak56 [21]

Answer:

51.0791%

Step-by-step explanation:

Calculate percentage change

from V1 = 41700 to V2 = 63000

(V2−V1)|V1|×100

=(63000−41700)|41700|×100

=2130041700×100

=0.510791×100

=51.0791%change

=51.0791%increase

7 0
3 years ago
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