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2 years ago

A truck weighs 72429 pounds.when rounded to the nearest ten-thousands the truck's weight will be

Mathematics

1 answer:

marta [7]2 years ago

6 0

I believe the answer is 70000

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Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

4 0

2 years ago

Emma is playing miniature golf. Her ball is 2 feet to the left of the right-hand wall. The hole is 4 feet to the left of the wal

NISA [10]

Answer:

She should aim 6 feet down the wall

Explanation:

The diagram attached sketches the situtation.

Since the angle with which the ball hits the wall is the same with which it bounces, angle β is the same for the two shown triangles.

Then, since both are right triangles, then all the angles are congruent and the triangles are similar. Hence, you can equal the ratios of the sides, to make an equation:

x/4 = y/2 ⇒ x = 2y

You have other equation:

x+ y = 18

Substitute

2y + y = 18
3y = 18
y = 18/3
y = 6 ← this is the distance down the wall where the ball should hit

x = 2(6) = 12

Then, she should aim 6 feet down the wall.

8 0

3 years ago

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

irga5000 [103]

Answer:

The rocket hits the ground at a time of 11.59 seconds.

Step-by-step explanation:

The height of the rocket, after x seconds, is given by the following equation:

$y = -16x^2 + 177x + 98$

It hits the ground when $y = 0$ , so we have to find x for which $y = 0$ , which is a quadratic equation.

Finding the roots of a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$y = -16x^2 + 177x + 98$

$-16x^2 + 177x + 98 = 0$

$a = -16, b = 177, c = 98$

$\bigtriangleup = 177^{2} - 4(-16)(98) = 37601$

$x_{1} = \frac{-177 + \sqrt{37601}}{2*(-16)} = -0.53$

$x_{2} = \frac{-177 - \sqrt{37601}}{2*(-16)} = 11.59$

Since time is a positive measure, the rocket hits the ground at a time of 11.59 seconds.

4 0

3 years ago

a drilling team detected water 38 feet below the ground. if it took workers 8 days to be able to dig and pull up the water, how

xz_007 [3.2K]

Answer:

On average, the workers dig 4.75ft each day. Also, what x and y questions?

Step-by-step explanation:

Take $38/8=4.75$

I hope you have a great day and make me the brainliest.

3 0

2 years ago

The triangles shown are congruent by the SSS congruence theorem.

denis23 [38]

1. Let's check the problem backwards. So let's map A"B"C" to ABC

2. We first need to reflect A"B"C" to A"B"C' with A"B"" as the axis of rotation.

3. Then we shift a few units up (translation) A"B"C' to A'B'C

4. Finally we rotate A'B'C around C to map the triangle onto ABC

going backwards we get the answer: b

b.rotation, then translation, then reflection

3 0

2 years ago

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