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3 years ago

Whats the answer to all of these

Mathematics

1 answer:

vova2212 [387]3 years ago

3 0

2,000 lb = 1 ton
25 ton = 50,000 lb
26 lb = 416 oz
24 ton = 48000 lb
2 ton = 4000 lb
19 ton = 38000 lb
12 lb = 192 oz
36 ton = 72000 lb
512 oz = 32 lb
40 lb = 640 oz
30 ton = 960000
34 ton = 1.088e+6
37 lb = 592 oz
38 ton = 76000

There are the answers! :)

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A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first the

Svetlanka [38]

Answer:

The hotel charge in each city before tax was $5125 of the first city and $3625 of the second city.

Step-by-step explanation:

Given:

A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first.

The tax of the first city was 6% and the tax of the second city was 10%.

Total hotel tax paid for two cities with $670.

Now, to find the hotel charge in each city before tax.

Let the hotel charge in first city before tax be $x.$

And the hotel charge in second city before tax be $y.$

So, as the hotel charge of the second city was 1500 higher than the first.

Thus,

$y=x-1500$ ........(1)

And as given, the tax of the first city was 6% and the tax of the second city was 10%, total hotel tax paid for two cities with $670.

6% of $x$ + 10% of $y$ = $670.

$\frac{6x}{100} +\frac{10y}{100} =670$

$0.06x+0.10y=670$

Substituting the value of $y$ from equation (1) we get:

$0.06x+0.10(x-1500)=670$

$0.06x+0.10x-150=670$

$0.16x-150=670$

Adding both sides by 150 we get:

$0.16x=820$

Dividing both sides by 0.16 we get:

$x=5125.$

The hotel charge in first city before tax = $5125.

Now, substituting the value of $x$ in equation (1) we get:

$y=x-1500$

$y=5125-1500$

$y=3625.$

The hotel charge in second city before tax = $3625.

Therefore, the hotel charge in each city before tax was $5125 of the first city and $3625 of the second city.

5 0

3 years ago

In an older television set, the screen height is 0.75 times the screen width.

natima [27]

The television set has a rectangular shape. The diagonal of this rectangle along with its width and length together form a right angular triangle.
This means that we can apply the Pythagorean theorem which states that:
(diagonal)^2 = (length)^2 + (width)^2

Let the width be w. We know that the length is 0.75 times the width, this means that: length = 0.75 w

Substitute in the above equation:
(20)^2 = (0.75w)^2 + (w)^2
400 = 0.5625 w^2 + w^2
400 = 1.5625 w^2
w^2 = 256
w = 16

This means that the width of the screen is 16 in.

5 0

3 years ago

Please answer, thanks! :)

irinina [24]

What’s the question that you have

7 0

2 years ago

Let f be the function defined by f(x) = e^(x) cos x.

Pavel [41]

(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

$\displaystyle
f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}$

____________

(b)

$f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\
f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\
f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}$

The slope of the tangent line is $e^{3\pi/2}$ .

____________

(c)

The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.

Solving f'(x) = 0

$f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\
0 = e^x \big( \cos(x) - \sin(x)\big)$

Use zero factor property to solve.

$e^x \ \textgreater \ 0\forall x \in \mathbb{R}$ so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

$\cos (x) - \sin (x) = 0 \\
\cos(x) = \sin(x)$

cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):

$\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\
x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]$

We check the values of f at the end points and these two critical numbers.

$f(0) = e^1 \cos(0) = 1$

$\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4} \frac{\sqrt{2}}{2}$

$\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4} \frac{-\sqrt{2}}{2} = -e^{\pi/4} \frac{\sqrt{2}}{2}$

$f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}$

There is only one negative number.
The absolute minimum value of f on the interval 0 ≤ x ≤ 2π is
$-e^{5\pi/4} \sqrt{2}/2$

____________

(d)

The function f is a continuous function as it is a product of two continuous functions. Therefore, $\lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0$

g is a differentiable function; therefore, it is a continuous function, which tells us $\lim_{x \to \pi/2} g(x) = g(\pi/2) = 0$ .

When we observe the limit $\displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}$ , the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)}$

$f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\
g'(\pi/2) = 2$

thus

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}$

3 0

3 years ago

Referring to the Fig. in Question #50, find the tangent of ∠A. Give answer in simplest form.

polet [3.4K]

Answer:

You're answer is 0.42.

Step-by-step explanation:

You first step is to find the angle of a which is 22.62.

Then you take the Tangent of 22.62 which it 0.42, hope this helps!

4 0

2 years ago

Read 2 more answers