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Simora [160]
1 year ago
9

Last year the girls’ basketball team had 8 fifth-grade students and 7 sixth-grade students. What was the ratio of sixth-grade st

udents to fifth-grade students on the team?
Mathematics
2 answers:
attashe74 [19]1 year ago
6 0

7:8 it could also be written as 7 to 8 or 7/8

Since we need to find out SIXTH to FIFTH grades we have to put the sixth graders first. Hope this helps and have a great day!

timofeeve [1]1 year ago
4 0

Answer:

the ratio of the sixth-grade students to fifth-grade students on the team is 7 : 8

Step-by-step explanation:

Given Parameters

Fifth grade students of the basket team = 8

Sixth grade students of the basket team = 7

Required;

Ratio of the sixth-grade students to fifth-grade students on the team

Let F represent the fifth graders

F = 8

Let S represent the sixth graders

S = 7

The ratio of S to F is represented mathematically by S : F

By substituton

S : F becomes

7 : 8

Hence, the ratio of the sixth-grade students to fifth-grade students on the team is 7 : 8

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Simplify 5 square root of 11 end root minus 12 square root of 11 end root minus 2 square root of 11 . (1 point)
Feliz [49]

Answer:-9√(11)

Step-by-step explanation:

5√(11) - 12√(11) - 2√(11)

Since they are all alike,as in they possess √(11),we can just add or subtract them through

-7√(11) - 2√(11)

-9√(11)

3 0
2 years ago
A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o
Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

\dfrac{dA}{dt}=R_{in}-R_{out}

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}

Change In Volume of the tank, \frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

<u>Rate Out</u>

Rate Out=(concentration of salt in outflow)(output rate of brine)

R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}

Therefore:

\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3

This is a linear differential equation in standard form, therefore the integrating factor:

e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2

Multiplying the DE by the integrating factor, we have:

(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000

Therefore, the amount of salt in the tank at any time t is:

A(t)=(50+t)-75000(50+t)^{-2}

After 15 minutes, the amount of salt in the tank is:

A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds

8 0
2 years ago
A carpenter is framing an opening in a wall to hang a rectangular door. Which properties must the door frame have to ensure that
tigry1 [53]
A) The top and bottom are parallel.B)The top and right side form a right angle.D)The left side and right side are parallel.E)The left side and bottom form a right angle.
Are correct just took the assignment and got it right.
5 0
2 years ago
Read 2 more answers
An expression is given: 2 open parentheses square root of k minus 1 close parenthesis plus square root of 8. If on adding negati
neonofarm [45]

Answer:

Possible value of k is √2

Step-by-step explanation:

The information given are;

The expression, 2·(√k - 1) + √8 to which may be added -6·√2 to obtain a rational number, we therefore have;

2·(√k - 1) + √8 - 6·√2 = R

Therefore, simplifying gives;

2·√k - 2 + 2·√2  - 6·√2 =  2·√k - 2 - 4·√2 = R

2·√k - 2 - 4·√2 + 2= R + 2 = R

2·√k - 2+ 2 - 4·√2 = R

2·√k - 2+ 2 - 4·√2 = R

2·√k + 0 - 4·√2 = 2·√k - 4·√2 = 2·(√k - 2·√2) = R

(√k - 2·√2) = R/2 = R

Therefore, √2 is a factor of √k such that √k - 2·√2 = R

Which gives k = x·√2, where x = a rational number

When x = 1, k = √2.

Therefore, a possible value of k is √2

3 0
2 years ago
Jake went on a riverboat tour. After returning from the tour, he was curious to know the speed at which the boat was going. Duri
atroni [7]
I'm sorry I'm too lazy to read all of that. However, I did get the main point. The river was going at 3 mph. Since he wanted to travel 10 miles upstream and downstream, you would have to subtract 20 by 3. This would mean he was going at a speed of 17 mph. 
4 0
2 years ago
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