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Setler [38]
2 years ago
12

The graph below shows the solution set of which inequality ? help guys plz ):

Mathematics
1 answer:
pickupchik [31]2 years ago
7 0
I'm pretty sure it's A
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If f(x)=sinx−1/cos2x, then limx→π2f(x) is equivalent to which of the following?
Harrizon [31]

Answer:

2

Step-by-step explanation:

Given g(x) = sin(x)-1/cos2(x), we are to find the limit if the function g(x) as g(x) tends to π/2

Substituting π/2 into the function

lim x-->π/2 sin(x)-1/cos 2(x)

= sin(π/2) - 1/cos(2)(π/2)

= 1 - 1/cosπ

= 1- 1/-1

= 1 -(-1)

= 1+1

= 2

Hence the limit of the function h(x) = sin(x)-1/cos2(x) as x--> π/2 is 2

8 0
2 years ago
Jason hires groundskeepers. They work an average of 60 hours per month. If seven groundskeepers are hired at $12/hour, what is t
Kamila [148]
Multiply the cost per hour by 60 to find the cost for a groundskeeper
cost = 12 × 60
cost = 720
The cost for each groundskeeper is $720

Find the cost for seven groundkeepers
Multiply cost for each groundskeeper by 7
cost = 720 × 7
cost = 5,040

The monthly labor cost for seven groundskeepers is $5,040
5 0
2 years ago
A popular podcast poses questions to subscribers at the end of each segment, asking listeners to enter their
mario62 [17]

Answer:

The poll uses voluntary response, so the results tell us little about the population of all adults.

Step-by-step explanation:

I took the test

8 0
2 years ago
Read 2 more answers
Write if you need to multiply or divide, then solve.
blsea [12.9K]

Answer:

Divide. Answer is 4

Step-by-step explanation:

First, divide.

235/75=3\frac{1}{3}

Since you can't have a third of a box, you need the next number up, which is 4.

3 0
2 years ago
A tank contains 8000 L of pure water. Brine that contains 35 g of salt per liter of water is pumped into the tank at a rate of 2
OleMash [197]

Answer:

The concentration of salt in the tank approaches 35 \mathrm{g} / \mathrm{L},

Step-by-step explanation:

Data provide in the question:

Water contained in the tank = 8000 L

Salt per litre contained in Brine = 35 g/L

Rate of pumping water into the tank = 25 L/min

Concentration of salt \lim _{t \rightarrow \infty} C(t)=\lim _{t \rightarrow \infty} \frac{35 t}{320+t}

Now,

Dividing both numerator and denominator by t, we have

\lim _{t \rightarrow \infty} \frac{\frac{1}{t} 35 t}{\frac{1}{t}(320+t)}=\lim _{t \rightarrow \infty} \frac{35}{\frac{320}{t}+1}=35

Here,

The concentration of salt in the tank approaches 35 \mathrm{g} / \mathrm{L},

3 0
2 years ago
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