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Neporo4naja [7]
3 years ago
13

How do you find x? Please help

Mathematics
1 answer:
marusya05 [52]3 years ago
5 0
Since you know that DA and DC are equal in length, you can say 2x-3=x+2, when you add 3 to both sides, it becomes 2x=x+5, subtract x from both sides, and you find that x=5
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Arlecino [84]
The listed ordered pairs that are solutions are D
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3 years ago
Is question a. Linear or non linear
Zanzabum

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yes

Step-by-step explanation:

5 0
3 years ago
help ASAP find the coordinates of the midpoint of the given segment given its endpoints are (-4,3) and (2,-1)​
PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

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7 0
3 years ago
A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of
liq [111]

Answer:

The probability of failure of both the bulbs  is 0.4323.

Step-by-step explanation:

For an exponential distribution the distribution is given by

f(x,\lambda )=\int_{0}^{x }\lambda e^{-\lambda x}dx

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

P(X)=\int_{0}^{x }\lambda _{X}e^{-\lambda _{X}x}dx

Similarly for the bulb Y the distribution function is given by

P(Y)=\int_{0}^{y }\lambda _{Y}e^{-\lambda _{Y}y}dy

Thus the probability for both the bulbs to fail within 1500 hours is

P(E)=\int_{0}^{1500}\int_{0}^{1500}\frac{1}{1400}e^{\frac{-x}{1400}}\cdot \frac{1}{1400}e^{\frac{-y}{1400}}dxdy\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}\int_{0}^{1500}e^{\frac{-x}{1400}}\cdot e^{\frac{-y}{1400}}dxdy)\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}e^{\frac{-x}{1400}}dx)\cdot (\int_{0}^{1500}e^{\frac{-y}{1400}}dy)\\\\P(E)=\frac{1}{1400^{2}}\times 920.473\times 920.473\\\\\therefore P(E)=0.4323

6 0
3 years ago
What is the value of g?
lyudmila [28]

Answer:

90

Step-by-step explanation:

I'm not sure about the answer

8 0
2 years ago
Read 2 more answers
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