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4 years ago

The equation tan^2 x+1=sec^2 x is an identity true or false

Mathematics

1 answer:

Solnce55 [7]4 years ago

8 0

Answer:

tan²x + 1 = sec²x is identity

Step-by-step explanation:

* Lets explain how to find this identity

∵ sin²x + cos²x = 1 ⇒ identity

- Divide both sides by cos²x

∵ sin x ÷ cos x = tan x

∴ sin²x ÷ cos²x = tan²x

- Lets find the second term

∵ cos²x ÷ cos²x = 1

- Remember that the inverse of cos x is sec x

∵ sec x = 1/cos x

∴ sec²x = 1/cos²x

- Lets write the equation

∴ tan²x + 1 = 1/cos²x

∵ 1/cos²x = sec²x

∴ than²x + 1 = sec²x

- So we use the first identity sin²x + cos²x = 1 to prove that

tan²x + 1 = sec²x

∴ tan²x + 1 = sec²x is identity

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Ymorist [56]

Answer:

1-point: 4
2-point: 9
3-point: 3

Step-by-step explanation:

Let x, y, z represent the numbers of 1-, 2-, and 3-point baskets, respectively. The given relations are ...

x +2y +3z = 31 . . . . . . total points scored

y -x = 5 . . . . . . . . . . . 5 more 2-point baskets than free throws

y = 3z . . . . . . . . . . . . 3 times as many 2-point baskets as 3-point baskets

Using the second equation to write an expression for x, we have ...

x = y -5

Substituting this and the third equation into the first, we get ...

(y -5) +2y +(y) = 31

4y = 36 . . . . . . . . . . . add 5, collect terms

y = 9

x = 9 -5 = 4

z = y/3 = 9/3 = 3

The player made 4 free throws, 9 2-point baskets, and 3 3-point baskets.

_____

Alternate solution

The equations can be written in augmented matrix form as ...

$\left[\begin{array}{ccc|c}1&2&3&31\\-1&1&0&5\\0&1&-3&0\end{array}\right]$

Row-reducing this matrix using any of a number of calculators gives (x, y, z) = (4, 9, 3), as above.

3 0

3 years ago

Read 2 more answers

2x^3=28 find the exact solution for x

Alborosie

Answer:

The answer to your question is: x = 2.41

Step-by-step explanation:

2x³ = 28

x³ = 28/2

x³ = 14

x = ∛14

x = 2.41

6 0

3 years ago

Whats 16+15 in factored form

VashaNatasha [74]

FULL ANSWERFactored form may be a product of greatest common factors or the difference of squares. For instance, the factored form of x^3 + 2x^2 - 6 = x(x+2)(x-3) and the factored form of x^2 - 16 = (x+4)(x-4)It is possible to solve for x by using factored form; x^2 + 5x + 6 can be reduced to its factored form by removing the x as a common factor. This results in (x+2)(x+3). Once in the factored form, solve for x by multiplying to get zero. In the equation (x+2)(x+3) = 0, the zero factor property explains that anything multiplied by zero equals zero. This means x + 2 = 0 and x + 3 = 0. The solutions to this formula would be x = -2 and x = -3.Once the solutions for x are found, check to make sure they work. In the equation x^2 + 5x + 6 = 0, it was found that x = -2, -3. Replace x in the equation with each of the solution values. So, [-2]^2 + 5(-2) + 6 = 0 turns into 4 - 10 + 6 = 0, which is correct, and [-3]^2 + 5(-3) + 6 = 0 becomes 9 - 15 + 6 = 0, which is also correct.
i hope this will help u w/ your question

7 0

4 years ago

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$