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VladimirAG [237]
3 years ago
8

It takes 3 hours for marty and cora to weed a garden. How long will it take 6 people to weed the same garden if they all work at

the same rate? Explain
Mathematics
1 answer:
klemol [59]3 years ago
5 0

Answer: 2 hours

Step-by-step explanation:

if you split it by 2 people each it would take two hours and 6 divided by 3 = 2

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Find last year’s salary if this year’s salary is $48,200 after a 3% pay raise ?
viktelen [127]

Answer:

49646

Step-by-step explanation:

6 0
3 years ago
Decide whether quadrilateral JKLM is a rectangle, a rhombus, or a square. More than one answer may apply!
NARA [144]

Answer:

Square.

Step-by-step explanation:

given J(-4,2) || K(0,3) || L(1,-1) || M(-3,-2)

given below the visual answer.

OR

you can find the distance of coordinate to coordinate:

Distance of JK

√(-4-0)² + (2-3)²

√17

Distance of KL

√(0-1)²+(3--1)²

√17

Distance of LM

√(1--3)²+(-1--2)²

√17

Distance of MJ

√(-3--4)²+(-2-2)²

√17

All sides are similar of √17 , so its a square.

4 0
2 years ago
Read 2 more answers
Plz help me solve this
posledela

1.-28×-28×-28+13×13×13+16×16×16

=21,952+2,197+4,096

=28,245.

2.12×12×12+-7×-7×-7+-5×-5×-5

=1,728+-343+-125

=1,260

6 0
3 years ago
What is the value of a, b, and c in the quadratic equation -2x^2+4x-3=0
amm1812
A = -2
b = 4
c = -3

They are the coefficients to the terms in the equation<span />
6 0
3 years ago
Read 2 more answers
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
3 years ago
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