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3 years ago

Secθ =__________________ 1/sinθ +-√cos^2 +1 +-√tan^2 +1

Mathematics

2 answers:

Zielflug [23.3K]3 years ago

6 0

Answer:

The answer is last option

As we know 1 + tan^2 = sec^2

denis-greek [22]3 years ago

5 0

Answer:

secФ = ± $\sqrt{1+tan^{2} }$

Step-by-step explanation:

We can use the identity sec²Ф = 1 + tan²Ф to solve for secФ

First we need to take the square root of both sides to get: $\sqrt{sec^{2} } =\sqrt{1+tan^{2} }$ which then simplifies to
secФ = ± $\sqrt{1+tan^{2} }$

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What statements about the graphs of these two functions are true?

tamaranim1 [39]

Answer:

B)Only one graph intersects the y-axis above y=0

Step-by-step explanation:

Both functions are in y=mx+b. The y-intercept is b.

f(x) has the y-intercept 3. It intersects above y=0.

g(x) has the y-intercept -4. It intersects below y=0.

This means B is correct.

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2 years ago

What are the two words that allow you to check your division problem backward?

Dahasolnce [82]

Answer:

inverse operation

Step-by-step explanation:

Inverse means opposite. The opposite of division is multiplication. For example, if you divided 15 by 3, the answer is 5. If you multiply 5 by 3, the answer is 15. You know your answer is correct by checking it with the inverse operation.

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3 years ago

The football team is raising money to have a new turf field installed the cost of the turf field is 48780 the team has 18 months

Stells [14]

I think the answer is 876744

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2 years ago

A radioactive substance has a continuous decay rate of 0.056 per minute. How many grams of a 120 gram sample will remain radioac

sammy [17]

Answer:

The mass of the radioactive sample after 40 minutes is 12.8 g.

Step-by-step explanation:

The mass of the sample can be found by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

N(t): is the amount of the sample at time t =?

N₀: is the initial quantity of the sample = 120 g

t = 40 min

λ: is the decay constant = 0.056 min⁻¹

Hence, the mass of the sample after 40 min is:

$N_{t} = N_{0}e^{-\lambda t} = 120g*e^{-0.056min^{-1}*40 min} = 12.8 g$

Therefore, the mass of the radioactive sample after 40 minutes is 12.8 g.

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3 years ago

1 + tanx / 1 + cotx =2

Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer: x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0

3 years ago