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uranmaximum [27]
3 years ago
11

Help me writing and solving the inequality for each problem.

Mathematics
1 answer:
MariettaO [177]3 years ago
5 0
They would need to sell 31 bottles
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Find an nth-degree polynomial function<br> n= 3;<br> - 2 and 6 + 2 i are zeros;<br> f(-1)= 53
Bas_tet [7]

Answer:

Find an nth-degree polynomial function.

Step-by-step explanation:

n= 3;

- 2 and 6 + 2 i are zeros;

f(-1)= 53

8 0
2 years ago
Justin and his children went into a bakery and where they sell cookies for $0.75 each and brownies for $1.25 each. Justin has $1
enot [183]

Answer:

Justin could buy up to 8 cookies.

Step-by-step explanation:

7 * 1.25 = 8.75

15 - 8.75 = 6.25

6.25/.75 = 8.3333333

5 0
3 years ago
Shanice's car is traveling 10 miles per hour slower than twice the speed of Brandon's car. She covers 93 miles in 1 hour 30 minu
joja [24]

Shanice's rate: 93 / 1.5 = 62 miles per hour

she is 10 miles slower than twice the speed of Brandon

62 +10 = 72

72/2 = 36

 Brandon is driving 36 miles per hour

3 0
3 years ago
If $10,000 is invested at a rate of 2.5 percent for 5 years, how much money is in the account at the end of the term?
Vladimir [108]
10,000(1 + .025 x 5)
11,250
4 0
3 years ago
Read 2 more answers
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

Where,

A_0 = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

i.e.

\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

\implies k = \frac{\ln(0.5)}{19}\approx -0.03648

Now, if the substance to decay to 78​% of its original​ amount,

Then A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0

0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0
2 years ago
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