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antiseptic1488 [7]
3 years ago
13

Please help with this

Mathematics
1 answer:
Elza [17]3 years ago
8 0

area \: of \: circle \: a1 = \pi {r}^{2}  = \pi \times {21}^{2}  = \\  441\pi sq \: \:  .in \\ area \: of \: triangle \: a2 =  \frac{1}{2} bh =  \frac{1}{2}  \times 20 \times 39  \\ = 390 \: sq.in \\ now \: area \: of \: shaded \: region = a1 - a2 \\  = 441\pi - 390 \\  = 995.44 \: sq.in

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. Un turist a parcurs un traseu în trei zile. În prima zi turistul a parcurs
Dahasolnce [82]

Answer:

The total length of route travelled in 3 days = 30 km

Lungimea traseului parcurs în cele trei zile = 30 km

Step-by-step explanation:

English Translation of the Question

A tourist traveled a route in three days. On the first day the tourist covered 40% of the length of the route, on the second day the tourist covered 5/6 of the remaining distance after the first day, and on the third day the remaining 3 km. Calculate the length of the route traveled in the three days.

Let the total length of the route be x.

Lungimea traseului parcurs în cele trei zile = x

- On the first day, the tourist covers 40% of the total length of the route;

În prima zi turistul a parcurs 40% din lungimea traseului

That is, 40% of x = 0.4x

- On the second day, the tourist covered 5/6 of the remaining journey.

în a doua zi turistul a parcurs 5/6 din distanța rămasă de parcurs după prima zi

The remaining journey after day 1 = x - 0.4x = 0.6x

(5/6) of the remaining journey = (5/6) × 0.6x = 0.5x

- On the third day, the toursit travels the total remaining length of the route = 3 km

iar în a treia zi restul de 3 km

The total remaining length of the route = x - 0.4x - 0.5x = 0.1x

0.1x = 3

x = (3/0.1) = 30 km

The Hope this Helps!!!

3 0
3 years ago
byron purchased a $5000 certificsye of deposit at his local bank the cd will pau him 7% simple interest at the end of 2 years ho
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5000(1.07)^2
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BARSIC [14]
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We have x^2-2x-3 in the denominator, so:

x^2-2x-3\neq0\\\\(x^2-2x+1)-4\neq0\\\\(x-1)^2-4\neq0\\\\(x-1)^2-2^2\neq0\qquad\qquad[\text{use }a^2-b^2=(a-b)(a+b)]\\\\(x-1-2)(x-1+2)\neq0\\\\
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So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.

2. Asymptotes:

f(x)=\dfrac{x-3}{x^2-2x-3}=\dfrac{x-3}{(x-3)(x+1)}=\boxed{\dfrac{1}{x+1}}

We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)
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9 and 10

11 and 12

13 and 14

15 and 16

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