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kap26 [50]
2 years ago
12

What fractions are equivalent to 28/32

Mathematics
2 answers:
marin [14]2 years ago
7 0
The answer is 7/8. 28/4 = 7, 32/4=8.
Jlenok [28]2 years ago
5 0
÷2==14/16
÷4==7/8

HOPE THIS HELPS
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Does the equation 2x-7=5+x have exactly one solution
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Answer:

12

Step-by-step explanation:

2x - 7 =5 +x

2x - x =5 +7

x =12 (one solution)

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Step-by-step explanation:

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C. None of the x-values repeat.
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A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

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so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
3 0
3 years ago
I will mark branliest for ever Andrews first and it’s also 40 points
navik [9.2K]

Answer:

see below

Step-by-step explanation:

5/10 divide ?/?= 1/2

5/10 divide by 5/5 = 1/2

2. 12/15 divide ?/? = 4/5

12/15 divide by 3/3 = 4/5

3. 10/12divide ?/? = 5/6

10/12 divide by 2/2 = 5/6

4. 14/18 divide ?/? = 7/9

14/18 divide by 2/2 = 7/9

7 0
3 years ago
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