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s344n2d4d5 [400]
3 years ago
9

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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3 years ago
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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).
Hope this helps.
3 0
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:

#Princesses Rule

6 0
2 years ago
A thermometer at the park shows a reading of 86° Fahrenheit.
Marina86 [1]
No he will not.
86 degrees F is 30 degrees C
5 0
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
4 0
2 years ago
Read 2 more answers
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