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galina1969 [7]
2 years ago
11

If a function has a verticale asymptote at a certain x-value, then the function is at the value

Mathematics
1 answer:
stiks02 [169]2 years ago
8 0

Answer:

We conclude that If a function has a vertical asymptote at a certain x-value, then the function is undefined at the value.

Step-by-step explanation:

If a function has a vertical asymptote at a certain x-value, then the function is undefined at the value.

For example, let the function

f\left(x\right)\:=\frac{x+3}{x-3}

It is clear that the given function becomes undefined at x = 3 in the denominator.

i.e. 3-3 = 0

It means, the function can not have x = 3, otherwise, the function will become undefined.

In other words, if the function has a vertical asymptote at x = 3, then the function is undefined at the value.

Therefore, we conclude that If a function has a vertical asymptote at a certain x-value, then the function is undefined at the value.

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shirley purchased a plot of land for $19,500. the land appreciates about 3.9% each year. what is the value of the land after 5 y
romanna [79]

Answer:

$23,302.50

Step-by-step explanation:

4 0
3 years ago
Rewrite in simplest terms: 2(-4p-6)-5p2(−4p−6)−5p
irina1246 [14]

Answer: 20p³+30p²-13p-12

Step-by-step explanation:

2(-4p-6)-5p²(-4p-6)-5p

-8p-12+20p³+30p²-5p

20p³+30p²-13p-12

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3 0
3 years ago
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A gold mine has two​ elevators, one for equipment and another for the miners. The equipment elevator descends 4 feet per second.
Ad libitum [116K]

Let d_m,\ d_e represent the depth of the miner and equipments elevator, respectively. We know that equipment elevator descends with a velocity of v_e = 4 feet per second, and the miners one descends with a velocity of v_m = 15 feet per second.

If we start counting time (t=0) when the equipment elevator begins to descend, after t seconds its depth will be

d_e(t) = v_et = 4t

On the other hand, the miner elevator follows the same rule, but we have to use its velocity, and remember that it starts with a delay of thirty seconds:

d_m(t) = v_m(t-30) = 15(t - 30)

Now, we have to wait the 30 seconds of delay, and then another 14 seconds. This means that we want to know the positions of both elevators when t = 44. Let's plug this value into the two equations:

d_e(44) = 4\cdot 44 = 176

d_m(44) = 15(44-30) = 15\cdot 14 = 210

So, the equipment elevator is 176 feet deep, and the miner elevator is 210 feet deep, and thus this is the deeper one.

7 0
3 years ago
You begin with $90 in your savings account and your friend begins
Colt1911 [192]

Answer:

F(f) = 15t + 35 represents the total amount of savings your friend would make in t weeks.

F(d) = 10t + 90 represents the total amount of saving you, darian, would make in t weeks.

When you graph the equations, plugging in different values for t, you can see that the graphs  intersect at (11,200). This means that at 11 weeks, both you and your friend have the same amount of money saved up, $200.  They will not have the same amount of money in 10 weeks.

7 0
3 years ago
Read 2 more answers
Help me asap! I will give you marks
Law Incorporation [45]

Recall the binomial theorem.

(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k

1. The binomial expansion of \left(1+\frac x3\right)^7 is

\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}

Observe that

k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x

k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2

When we multiply these by 8-9x,

• 8 and \frac73 x^2 combine to make \frac{56}3 x^2

• -9x and \frac73 x combine to make -\frac{63}3 x^2 = -21x^2

and the sum of these terms is

\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}

2. The binomial expansion is

\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k

We get the a^6b^2 term when k=2 :

k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2

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