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il63 [147K]
3 years ago
10

The segments shown below could form a triangle. A. True B. False

Mathematics
1 answer:
Whitepunk [10]3 years ago
8 0
Very true. A triangle consists of three sides so think of how u could make a triangle with those three lines
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Rewrite 1/100,000,000 as a power of 10
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1/100,000,000 , note that 100,000,000 = 10⁸ , then:

1/100,000,000  = 1/10⁸
Remember that 1/aⁿ  = a⁻ⁿ

Hence 1/100,000,000 = 1/10⁸ = 10⁻⁸

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This is line segments
lianna [129]

Step-by-step explanation:

here \: given \: that \: total \: lenth = bd = 67 \\ then \: bd = bc + cd \\ so \: we \: can \: write \: that \\ 67 = 3x - 2 + 4x + 13 \\67 =  3x + 4x - 2 + 13 \\ 67 = 7x + 11 \\ 7x = 67 - 11 \\ 7x = 56 \\ x =  \frac{56}{7}  \\ x = 8 \\ thankyou

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How many pairs of potential clients can be randomly chosen from the pool of eight candidates?
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1 12 13 14 15 16 17 18
2 23 24 25 26 27 28
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6 67 68
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approximately 28 pairs. (breakdown above) hope this helps

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3 years ago
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What is the exponent of 50
german
2×5^{2} if you don't understand its 2 times 5 to the second power.
6 0
3 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
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