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Marat540 [252]
3 years ago
11

What multiples to 7 and adds to 22?

Mathematics
1 answer:
marshall27 [118]3 years ago
8 0
<span>. multiply to -18 add to -17 = Multiplying -18 and 1 will give -18 as the result and then on adding -18 and + 1 the result comes to -17
</span><span>2. multiply to 36 add to -13 = Multiplying -9 and -4 will give -36 as the result and then on adding -9 and -4 the result comes to -13
</span><span>3. multiply to -24 add to -5<span>= </span></span><span>Multiplying -8 and +3 will give -24 as the result and then on adding -8 and +3 the result comes to -5
4. multiply to -18 add to 7= </span><span>Multiplying +9 and -2 will give -18 as the result and then on adding +9 and -2 the result comes to 7 
5. multiply to -36 add to 9= </span><span><span>Multiplying +12 and -3 will give -36 as the result and then on adding +12 and -3 the result comes to 9
</span> 6. multiply to 24 add to 10= </span><span><span>Multiplying +6 and +4 will give 24 as the result and then on adding +6 and +4 the result comes to 10
</span> 7. multiply to 18 add to -9= </span><span><span><span>Multiplying -6 and -3 will give -18 as the result and then on adding -6 and -3 the result comes to -9
</span> 8. multiply to -36 add to -16</span>= </span><span>Multiplying -18 and +2 will give -36 as the result and then on adding -18 and +2 the result comes to -16</span>
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Brilliant_brown [7]
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7 0
3 years ago
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7. These number share the same percentage
otez555 [7]
Answer: 30

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24x 0.25= 6
6+24=£30

(Or 24x1.25=30)
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HELP ME WITH MATH ANGLES PLEASE BRAINLIST
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the second slide is A. 22

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6 0
3 years ago
A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min
Dmitry [639]

At any time t (min), the volume of solution in the tank is

500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}

If A(t) is the amount of salt in the tank at any time t, then the solution has a concentration of \dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}.

The net rate of change of the amount of salt in the solution, A'(t), is the difference between the amount flowing in and the amount getting pumped out:

A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)

Dropping the units and simplifying, we get the linear ODE

A'=10+5\sin\dfrac t4-\dfrac A{10}

10A'+A=100+50\sin\dfrac t4

Multiplying both sides by e^{10t} allows us to identify the left side as a derivative of a product:

10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}

\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}

e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt

Integrate and divide both sides by e^{10t} to get

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}

The tanks starts off with 30 lb of salt, so A(0)=30 and we can solve for C to get a particular solution of

A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}

6 0
3 years ago
During the election for class president,
Y_Kistochka [10]

Answer:

100

Step-by-step explanation:

.6*500=300, .4*500=200

300-200=100

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3 years ago
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