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Natalija [7]
3 years ago
13

Show the original equation, then solve for x

Mathematics
1 answer:
Lina20 [59]3 years ago
6 0

Answer: 24.7

cos 54° = x/42

⇔ x = 42. cos 54°

⇔ x ≈ 24.7

Step-by-step explanation:

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Romashka-Z-Leto [24]

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Step-by-step explanation:

<h3>22+2²=22+64=86</h3>
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Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum
lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\

\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\  (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\

=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3}  \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\

4 0
2 years ago
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Naily [24]

Answer: We know:

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This may sound obv, but it's important: garden A has 1 less side then garden B, which is why it can have longer sides but the same perimeter.

So side A1 + side A2 + side A3 = side B1 + side B2 +  side B3 + side B4.

A1 + A2 + A3 = B1 + B2 + B3 + 15(the extra length on the triangle sides)

Since B is a square, B = 15 m by 15 m.

I don't know enough about triangle to put the dementions, but it's an equilateral triangle so the sides are 20 m each.  Hope that helps! :)

3 0
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let's see...

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