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3 years ago

Which of the following is an extraneous solution of StartRoot negative 3 x minus 2 EndRoot = x + 2?

Mathematics

2 answers:

Studentka2010 [4]3 years ago

7 0

Answer:

x=-6

Step-by-step explanation:

deff fn [24]3 years ago

6 0

Answer:

x=-6

Step-by-step explanation:

just took the test

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Which of the following is the graph of (x + 1)2 + (y − 6)2 = 36?

Brrunno [24]

Answer:

Step-by-step explanation:

Slope:

− 1

y-intercept:

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3 years ago

Read 2 more answers

X f(x)

fgiga [73]

B) f(x) = 3

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3 years ago

a circular frame that is 3 inches wide surrounds the mirror what is the combined area, in square inches, of the circular mirror

NARA [144]

The answer is 2.25 square inches or 7.065 square inches.

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4 years ago

Carly's family traveled 3/10 of the distance to her grandmother's house on Saturday. They traveled 3/7 of the remaining distance

polet [3.4K]

Answer:

3/10 of the total distance to her grandmother's house was traveled on Sunday

Step-by-step explanation:

Carly's family traveled distance of Saturday = $\frac{3}{10}$

Remaining distance = $1-\frac{3}{10}=\frac{7}{10}$

They traveled 3/7 of the remaining distance on Sunday.

So, Distance traveled on Sunday = $\frac{3}{7}(\frac{7}{10})$

Distance traveled on Sunday = $\frac{3}{10}$

Fraction of the total distance to her grandmother's house was traveled on Sunday = $\frac{3}{10}$

Hence 3/10 of the total distance to her grandmother's house was traveled on Sunday

5 0

4 years ago

Does anybody know how to do time with exponential decay

My name is Ann [436]

<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

$b_n=0.12\times b_{n-1}$

Why does this work? Initially, you start with 500 mL of air that you breathe in, so $b_1=500\text{ mL}$ . After the second breath, you have 12% of the original air left in your lungs, or $b_2=0.12\timesb_1=0.12\times500=60\text{ mL}$ . After the third breath, you have $b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}$ , and so on.

You can find the amount of original air left in your lungs after $n$ breaths by solving for $b_n$ explicitly. This isn't too hard:

$b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots$

and so on. The pattern is such that you arrive at

$b_n=0.12^{n-1}b_1$

and so the amount of air remaining after $50$ breaths is

$b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}$

which is a very small number close to zero.</span>

5 0

3 years ago