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3 years ago

How to solve -2.4 × (-3.2)=

Mathematics

2 answers:

Zinaida [17]3 years ago

8 0

The answer is 7.68.

Hope this helps.

Fittoniya [83]3 years ago

7 0

7.68 because to neg make a pos

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Find x and y in the figure

xxTIMURxx [149]

Answer:

therefore angle Y will be equal to angle X by vertically opposite angle as by taking all the angles inside the traingle will be equal to Y so Y = X = 60°

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3 years ago

I WILL GIVE YOU BRAINLIEST ,5 STARS⭐, AND THANKS❤️. (P.S. AT LEAST TWO PEOPLE GIVE ANSWER SO I CAN GIVE BRAINLIEST)

Crank

Answer:

0.002

Step-by-step explanation:

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3 years ago

Math

Snezhnost [94]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the solution ~

$\qquad \sf \dashrightarrow \: 7.2 + (2x + 0.4)$

$\qquad \sf \dashrightarrow \: 7.2 + 2x + 0.4$

$\qquad \sf \dashrightarrow \: 7.2 +0.4 + 2x$

$\qquad \sf \dashrightarrow \: (7.2 +0.4) + 2x$

Therefore, the correct choice is A

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2 years ago

Read 2 more answers

In the right triangle shown, Angle A = 30° and AC = 18.

Komok [63]

Answer:

Step-by-step explanation:

Angle A plus AC and that adds up too 48

8 0

3 years ago

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

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3 years ago