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2 years ago

Determine the number of solutions of the equation

Mathematics

1 answer:

docker41 [41]2 years ago

6 0

I think there are only one possible answer

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Pls help ill give branliest

Mekhanik [1.2K]

Square- 1 1/2
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8 0

2 years ago

Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)

andrezito [222]

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$ , $0\leq x\leq 2\pi$ is when x = $\pi$

3 0

2 years ago

1. If a = -1 , the value of3a + 1 is

erastovalidia [21]

Answer:

$f(a) = 3a + 1 \\ Now,\: a = - 1 \\ f( - 1) = 3( - 1) + 1 \\ f( - 1) = - 3 + 1 \\ \boxed{ f( - 1) = - 2}$

Therefore, <em><u>-2</u></em> is the right answer.

6 0

1 year ago

14.

hoa [83]

D
You subtract 5 1/4 (5.25) - 3 3/4 (3.75) = 1 1/2 (1.5)

3 0

2 years ago

Read 2 more answers

Show that the sum of two concave functions is concave. Is the product of two concave functions also concave?

spayn [35]

Answer with explanation:

Let us assume that the 2 functions are:

1) f(x)

2) g(x)

Now by definition of concave function we have the first derivative of the function should be strictly decreasing thus for the above 2 function we conclude that

$\frac{d}{dx}\cdot f(x)$

Now the sum of the 2 functions is shown below

$y=f(x)+g(x)$

Diffrentiating both sides with respect to 'x' we get

$\frac{dy}{dx}=\frac{d}{dx}\cdot f(x)+\frac{d}{dx}\cdot g(x)\\\\$

Since each term in the right of the above equation is negative thus we conclude that their sum is also negative thus

$\frac{dy}{dx}$

Thus the sum of the 2 functions is also a concave function.

Part 2)

The product of the 2 functions is shown below

$h=f(x)\cdot g(x)$

Diffrentiating both sides with respect to 'x' we get

$h'=\frac{d}{dx}\cdot (f(x)\cdot g(x))\\\\h'=g(x)f'(x)+f(x)g'(x)$

Now we can see the sign of the terms on the right hand side depend on the signs of the function's themselves hence we remain inconclusive about the sign of the product as a whole. Thus the product can be concave or convex.

8 0

2 years ago