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Advocard [28]
2 years ago
7

If the height is 24, solve for x. h=-16x^2+8x+48

Mathematics
1 answer:
disa [49]2 years ago
8 0

Answer:

x=  

32

7+  

3121

​  

 

​  

,  

32

7−  

3121

​  

 

​  

 

Step-by-step explanation:

h+16^2-8x-48=0

-7x+16x^2-48=0

x=  

32

7+  

3121

​  

 

​  

,  

32

7−  

3121

​  

 

​  

 

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Solve z ÷ (–7) = –1.<br> A. 7 B. -7 C. 1/7 D. 6
aalyn [17]

Answer:

A. 7

Step-by-step explanation:

z ÷ (–7) = –1.

multiply each side by -7

z ÷ (–7)*-7 = –1.*-7

z = 7

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3 years ago
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I don't really grab the concept of this but it's basically numbering the graph ang filling the coordinates in the x and y spaces
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Pavel [41]

Answer:

The relationship is that if you divide say, 88÷4=22 the relationship is that all of them divided would equal 22.

Hope this helped.

3 0
2 years ago
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M is between L and N. If LN = 91 and LM = 23, find MN.
Tasya [4]

Answer:

MN = 68

Step-by-step explanation:

LN = 91

LM = 23

Points L, M, and N are collinear, therefore, according to the segment addition postulate, the following can be deduced:

LM + MN = LN

23 + MN = 91 (Substitution)

Subtract 23 from both sides

23 + MN - 23 = 91 - 23

MN = 68

8 0
2 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

5 0
3 years ago
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