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ANTONII [103]
4 years ago
10

Simplify this polynomial:3x2 - x + 2 - 5x2 + 8x - 5​

Mathematics
1 answer:
Lemur [1.5K]4 years ago
3 0

Answer:

3x - 3

Step-by-step explanation:

You might be interested in
Solve for x 11x+4<15 OR 12x−7>−25
never [62]

Isolate the x in both cases. What you do to one side, you do to the other.


11x + 4 < 15


Subtract 4 from both sides


11x + 4 (-4) < 15 (-4)


11x < 11


isolate the x, divide 11 from both sides


11x/11 < 11/11


x< 1


x < 1 is your answer for the first one

-------------------------------------------------------------------------------------------------------------------


Again, isolate the x.


12x - 7 > -25


Add 7 to both sides


12x - 7 (+7) > -25 (+7)


12x > -18


Isolate the x, divide 12 from both sides


12x/12 > -18/12


x > -1.5 is your answer for the second one.


-------------------------------------------------------------------------------------------------------------------



hope this helps

3 0
3 years ago
Read 2 more answers
Three point zero one divided by seven
vivado [14]
The answer is zero point four three

You could also write this answer as 0.43
7 0
3 years ago
Which of the following is equivalent to (-2/3) with a exponent of 3
Zinaida [17]

Answer:

( -8,27) I believe

Step-by-step explanation:

Because you times 2x2x2=8 and 3x3x3=27. Getting( -8,27)

Hope this helps :)

4 0
3 years ago
Uninhibited growth can be modeled by exponential functions other than​ A(t) ​=Upper A 0 e Superscript kt. For ​ example, if an i
laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

3 0
3 years ago
8. William got scores of q1, q2, q,3 on three quizzes.
Galina-37 [17]

Step-by-step explanation:

Average mean = (q1 + q2 + q3)/3.

7 0
3 years ago
Read 2 more answers
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