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k0ka [10]
2 years ago
12

Please help!!!!!!

Mathematics
1 answer:
elixir [45]2 years ago
4 0

Answer: A

Step-by-step explanation:

The area of the square is 10^2 = 100.

The area of the circle is (pi)(5^2) = 25pi.

So, the area of the region in the square but not in the circle is 100 - 25pi.

Thus, the desired probability is (100-25pi)/100, which is about 21.5%

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Write the limit as a definite integral on the interval [a, b], where ci is any point in the ith subinterval. Limit Interval lim
geniusboy [140]

Answer:

The corresponding definite integral may be written as

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x

The answer of the above definite integral is

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x = 98

Step-by-step explanation:

The given limit interval is

\lim_{||\Delta|| \to 0}  \sum\limits_{i=1}^n (4c_i + 11) \Delta x_i

[a, b] =  [-8, 6]

The corresponding definite integral may be written as

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x

\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x

Bonus:

The definite integral may be solved as

\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x \\\\\frac{4x^2}{2} + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\2x^2 + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\ 2(6^2 -(-8)^2 ) + 11(6 - (-8) \\\\2(36 - 64 ) + 11(6 + 8) \\\\2(-28 ) + 11(14) \\\\-56 +154 \\\\98

Therefore, the answer to the integral is

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x  = 98

5 0
3 years ago
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7 0
3 years ago
Wilson bought fruit that weighs pounds. How many ounces does the fruit weigh? Show your work
mrs_skeptik [129]
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3 years ago
Consider M, N, and P. collinear points on MP.
Kobotan [32]

Answer:

See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}

2. Point N lies to the right from point P, then MP + PN = MN.  Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}

3 0
3 years ago
Please help me I will give you bring me if it’s right
Otrada [13]

Answer:x

Step-by-step explanation:

7 0
2 years ago
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