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2 years ago

Sammy factored the polynomial x2 + x - 6. She came up with the factors of

Mathematics

1 answer:

shepuryov [24]2 years ago

5 0

Answer:

(x+3) and (x-2) are the correct factors

Step-by-step explanation:

If we expand (x-3)(x-2), we see that it equates to x²-5x+6, which is not what we want. The correct factors would be (x+3)(x-2) since the middle term's coefficient becomes 1 as 3-2=1 whereas -3-2=-5.

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3.52 in expanded form

Scrat [10]

Step-by-step explanation:

3.52 =

+ 0.5

+ 0.02

hopes this helps you

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2 years ago

In ΔKLM, k = 480 cm, ∠M=80° and ∠K=39°. Find the length of m

GuDViN [60]

Answer: 751

Step-by-step explanation:

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2 years ago

Read 2 more answers

A bag contains marbles numbered 1 through 100. Zoe draws a marble. Then, without replacing the marble, she draws a second marble

Neporo4naja [7]

If starting out, there are 100 marbles and Zoe pulls out a 5 and doesn't replace it, then there are 99 marbles left to choose from. So, there are 9 digits 1-10 (not including #5) making the proportion you have left 9/99 which simplifies to 1/11

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3 years ago

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$

d) $t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)$

$N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h$

6 0

3 years ago

For what values of t on the interval [0, 21 ]is the instantaneous velocity positive (the projectile moves upward)?

OLEGan [10]

Answer:

0 <=t<=21

Step-by-step explanation:

Projectile is Moving upwards on an interval of (0 to 21), if we plot Velocity vs Time and denote positive y-axis above 0 and negative y-axis below 0(for velocity), then from 0 to 21 t projectile is moving upwards and has positive velocity, when the projectile reaches the top of it's motion and returns back down to ground it's velocity is negative and is plotted below the y =0 (note that is for t > 21).

hence for the interval 0 <=t <=21 the instantaneous velocity is positive (Note, instantaneous velocity is also the derivative of the velocity or the slope ).

8 0

2 years ago