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Westkost [7]
3 years ago
10

How many t-shirts were sold if you spent a total of $88

Mathematics
1 answer:
torisob [31]3 years ago
5 0

Answer:

4 t-shirts because if one t-shirt is $22 then $88 divided by $22 is 4

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Integral 1+cos8x/tan2x-cot2x
UkoKoshka [18]
I believe it's<span> 8cos(x)⁸ - 16cos(x)⁶ + 10cos(x)⁴ - 2cos(x)².

</span>8cosx^8 - 16cosx^6+10cosx^4-2cosx^2<span>

Alternately, you can write [</span><span><span>1 / (tan(2x) - cot(2x))] + [cos(8x) / (tan(2x) - cot(2x))].

</span></span>\dfrac{1}{tan(2x)-cot(2x)}+ \dfrac{cos(8x)}{tan(2x)-cot(2x)}<span><span>
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5 0
3 years ago
A farmer has a truckload of watermelons to take to market. At the beginning of the trip, the watermelons weigh 1000 kg. 99% of t
Mariulka [41]

Answer:

The weight of watermelon at the end of the trip is <u>500 kg.</u>

Step-by-step explanation:

Given:

The weight of watermelons is 1000kg.

At the beginning of the trip watermelon is 99% of the weight:

99% of 1000kg

\frac{99}{100}\times 1000

=0.99\times 1000

= 990 kg

Now, at the end of the trip it is 98% of weight. So, let x kg of water has been evaporated. So, remaining weight of the watermelon is 1000-x.

If x kg of water is removed, then the amount of water left is 990-x.

As per question, water now weighs 98% of the total weight. So,

98% of (1000-x) = 990-x

\frac{98}{100}\times (1000-x)=990-x

980-0.98x=990-x

-0.98x+x=990-980 (Bringing like terms on one side)

0.02x=10\\x=\frac{10}{0.02}=500\ kg

Therefore, the weight of watermelon at the end of the trip is given as:

Weight = 1000 - x = 1000 - 500 = 500 kg

Therefore, the watermelon weighs 500 kg at the end of trip.

7 0
3 years ago
Read 2 more answers
Someone help please
Alla [95]

Answer:  Choice A

\tan(\alpha)*\cot^2(\alpha)\\\\

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

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

From this, we can say,

\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

Therefore, \tan(\alpha)*\cot^2(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}\\\\ is an identity. In mathematics, an identity is when both sides are the same thing for any allowed input in the domain.

You can visually confirm that \tan(\alpha)*\cot^2(\alpha)\\\\ is the same as \frac{\cos(\alpha)}{\sin(\alpha)}\\\\ by graphing each function (use x instead of alpha). You should note that both curves use the exact same set of points to form them. In other words, one curve is perfectly on top of the other. I recommend making the curves different colors so you can distinguish them a bit better.

6 0
2 years ago
1354 to the nearest ten
dusya [7]

Answer:

1350

Step-by-step explanation:

5 0
2 years ago
-8-6(1+7a)=32+4a
Sophie [7]

Answer:

b

Step-by-step explanation:

3 0
3 years ago
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