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PtichkaEL [24]
2 years ago
13

The value of X2 + 6x + 9 for x =4 is​

Mathematics
2 answers:
klasskru [66]2 years ago
7 0

Answer:

41

Step-by-step explanation:

putting x into the eqn

Natalija [7]2 years ago
3 0
41
4x2+6x4+9
You might want to double check my math
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Find the quotient.   22x 2 y 2 ÷ 11x 2 y 2
zmey [24]
<span>
22x^2y^2:11x^2y^2=\\\\ =\frac{22x^2y^2}{11x^2y^2}\\\\ =\frac{2(11x^2y^2)}{11x^2y^2}\\\\ =2</span>
8 0
3 years ago
Select the correct interpretation of the probability of getting an 11 when a pair of dice is rolled. Interpret an event as signi
JulijaS [17]

Answer:

1/18

Step-by-step explanation:

We are considering that we have 2 dices with 6 faces each (so, the probability to gettig any face in any dish is 1/6). To get an 11, we only have two ways to obtain it:

Dice 1= 6 and Dice 2 =5

or

Dice 1= 5 and Dice 2 =6

So, the probability of the event is given as:

P(Dice1=5 ∧ Dice2=6) ∪ P(Dice1=6 ∧ Dice2=5) = P(Dice1=5) x P(Dice2=6) + P(Dice1=6) x P(Dice2=5) = 1/6 x 1/6 + 1/6 x 1/6 = 1/36 + 1/36 = 2/36 = 1/18.

As 1/18 = 0,055, and 0,055 > 0,05, we consider the event as not significative (according to the definition of significance in the sentence).

7 0
3 years ago
Read 2 more answers
A hose fills a hot tub at a rate of 2.65 gallons per minute. How many hours will it take to fill a 283 ​-gallon hot​ tub?
dsp73

Answer:

you need to substitute values on the next form of resolving this problem as  follows...

A hose fills up a hot tub at a rate of 3.2 gallons per minute. How many hours will it take to fill a 300 gallon hot tub?

please explain the method of unit conversions as thoroughly as possible.

 

Solution:

 

The rate of fill up is, Rate = 3.2 Gallons / minute = 3.2 g/min

The hut tub volume is 300 Gallons

 

You can set up this problem as follows:

 

Every 3.2 gallons require 1 minute, How many minutes 300 gallons require?

 

3.2  g    1min

300 g     ? min    = (300 gallons x 1min/ 3.2 gallons)=(300/32)min

 

= 93.75 min

 

or simply the number minutes is the time required (T) the rate is (R) and the volume is (V)

such that T=V/R= (300g/3.2 g/min)= 93.75 min

7 0
3 years ago
Lanie's room is in the shape of a parallelogram. The floor of her room is shown below and has an area of 108 square feet. Lanie
dsp73

Answer:

Yes it would

Step-by-step explanation:

Lanie's room is in the shape of a parallelogram.

Lanie has a rectangular rug that is 6 feet wide and 10 feet long.

Area of a rectangle = Length × Width

Area of the rectangular rug = 10 feet × 6 feet

= 60 square feet

We are told that:

The floor of her room is shown below and has an area of 108 square feet.

Hence, the rug would fit on the floor of her room because it's area is within the area of the floor of her room.

7 0
2 years ago
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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