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3 years ago

Please help me!!!!! I need help!!!!!!!!!

Mathematics

1 answer:

Alex3 years ago

6 0

The answer is 145 divided by tan 65 degrees.

Since the equation for tangent is opposite/adjacent, you can just plug the numbers in.

tan(65)=145/x (x being the height that you are trying to find)

tan(65) * x = 145

Then divide tan(65) to isolate the x and you get:

145/tan(65)

Hope this helps!

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A drive-in theater has 20 rows of parking spaces with 15 spaces in each row. About how many parking spaces are there altogether?

Alex17521 [72]

300 parking spaces

Multiply the rows by the spaces in each row. 20 x 15 = 300

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What is the function

Ede4ka [16]

The graph shows us that the slope of f(x) is -2. Now we gotta find the slope of g(x) to compare it to that of f(x). The equation of g(x) is in slope-intercept form (y = mx + b, where m is the slope and b is the y-intercept), so the slope is given to us for that one as well: it's -6. A line with a slope of -6 will be steeper than a line with a slope of -3, therefore the answer is B - the slope of f(x) is less than the slope of g(x).
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3 years ago

Explain how to find the pre-tax cost of an item if you know the final cost and the sales tax rate.

JulijaS [17]

Answer:

Calculating total cost.Multiply the cost of the item or service by the sales tax in order to find out the total cost.The equation looks like this:Item or service cost x sales tax (in decimal form) = total sales tax.Add the total sales tax to the item or service cost to get your total cost.

Step-by-step explanation:

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2 years ago

A thermometer at the park shows a reading of 86° Fahrenheit.

Marina86 [1]

No he will not.
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3 years ago

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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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2 years ago

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