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Yanka [14]
3 years ago
8

PLEASE HELP THIS IS SUPER URGENT

Mathematics
1 answer:
vaieri [72.5K]3 years ago
8 0

Answer:

have no idea lol im terrible at math

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Find area of the figure
kirza4 [7]

Answer:

70 sq. in

Step-by-step explanation:

Area of the figure

= ( 10 x 5 ) + ( 5 x 4 )

= 50 + 20

= 70

= 70 sq. in

7 0
3 years ago
Twelve more than the product of a number and four is fewer than 60.
valina [46]
Product of a number and four is 4x 
<span>12 more than that is 12+4x </span>
<span>fewer than 60  </span>

<span>12 + 4x < 60</span>
7 0
3 years ago
Can someone help me with thiis?
3241004551 [841]
I believe the expression would be 7(g)+7
4 0
3 years ago
Mr. Tragrr has $500.00 to spend at a bicycle store. All the prices listed down below include tax.
fgiga [73]

To answer the question, you need to determine the amount Mr. Traeger has left to spend, then find the maximum number of outfits that will cost less than that remaining amount.

Spent so far:

... 273.98 + 3×7.23 +42.36 = 338.03

Remaining available funds:

... 500.00 -338.03 = 161.97

The cycling outfits are about $80 (slightly less), and this amount is about $160 (slightly more), which is 2 × $80.

Mr. Traeger can buy two (2) cycling outfits with the remaining money.

_____

The remaining money is 161.97/78.12 = 2.0733 times the cost of a cycling outfit. We're sure he has no interest in purchasing a fraction of an outfit, so he can afford to buy 2 outfits.

8 0
3 years ago
Read 2 more answers
If
likoan [24]

It's easy to show that 7\tan(4x) is strictly increasing on x\in\left[0,\frac\pi8\right]. This means

M = \max \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/12} = 7\sqrt3

and

m = \min \left\{7\tan(4x) \mid \dfrac\pi{16} \le x \le \dfrac\pi{12}\right\} = 7\tan(4x) \bigg|_{x=\pi/16} = 7

Then the integral is bounded by

\displaystyle 7\left(\frac\pi{12} - \frac\pi{16}\right) \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le 7\sqrt3 \left(\frac\pi{12} - \frac\pi{16}\right)

\implies \displaystyle \boxed{\frac{7\pi}{48}} \le \int_{\pi/16}^{\pi/12} 7\tan(4x) \, dx \le \boxed{\frac{7\sqrt3\,\pi}{48}}

7 0
2 years ago
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