All categories

2 years ago

Could someone please explain and answer these? Thank you!

Mathematics

2 answers:

MissTica2 years ago

7 0

So first thing more u need to add all the numbers and look at it and think. I pretty sure I'm right

Trava [24]2 years ago

6 0

Answer: C. 782,000(1.05)10

And

C.210

You might be interested in

two investments earn an annual income of $645. an investment earns an annual simple interest rate of 7.7%, and the other investm

Andreas93 [3]

In how many months or years if just one month a little under 3000 each

6 0

2 years ago

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

======================================================

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

2 years ago

The graph of the function f(x) = (x +2)(x + 6) is shown

dalvyx [7]

The true about the domain and the range of the function is:

The domain is all real numbers, and the range is all real numbers

greater than or equal to -4 ⇒ 1st answer

Step-by-step explanation:

f(x) = (x + 2)(x + 6) is a quadratic function with 2 factors (x + 2) and (x + 6)

By multiplying its two factors we will find the form of the quadratic function

∵ (x)(x) = x²

∵ (x)(6) = 6x

∵ (2)(x) = 2x

∵ (2)(6) = 12

∴ f(x) = x² + 6x + 2x + 12

- By adding like terms

∴ f(x) = x² + 8x + 12

The quadratic function represented graphically by a parabola

Look to the attached figure

The x-coordinate of the vertex point of the parabola h = $\frac{-b}{a}$

where b is the coefficient of x and a is the coefficient of x²

∵ f(x) = x² + 8x + 12

∴ a = 1 and b = 8

∴ h = $\frac{-8}{2*1}=-4$

The y-coordinate of the vertex point is k = f(h)

∵ h = -4

∴ k = f(-4)

∴ k = (-4)² + 8(-4) = 12 = 16 - 32 + 12

∴ k = -4

∴ The vertex point of the parabola is (-4 , -4)

∵ The parabola is opened upward

∴ Its vertex is minimum point

∴ The minimum value of f(x) is y = -4

∵ The domain of the function is the values of x

∵ The range of the function is the values of y corresponding to the

values of x

∵ x can be any real numbers

∴ x ∈ R, where R is the set of real numbers

∴ The domain of f(x) is all real numbers

∵ The minimum value of f(x) is y = -4

∴ y can be any real number greater than or equal to -4

∴ y ≥ -4

∴ The range is all real number greater than or equal to -4

The true about the domain and the range of the function is:

The domain is all real numbers, and the range is all real numbers

greater than or equal to -4

Learn more:

You can learn more about quadratic function in brainly.com/question/1332667

#LearnwithBrainly

7 0

2 years ago

What is 461/3261????

Gnoma [55]

The answer is 7.07 times, with rounding, still 7.07.

8 0

3 years ago

X2+8x=-1 complete the square

Sophie [7]

(X+4)^2-15

This is the answer

6 0

2 years ago