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Luba_88 [7]
3 years ago
11

What is the equation of the line that passes through the point (3,5) and is perpendicular to the line x=4

Mathematics
1 answer:
Mamont248 [21]3 years ago
7 0
Hello : 
an equation is : y-5 = 0(x-3)....<span> perpendicular to the line x=4 ( parrallel to x-axis)
so ; 
</span>y=5
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Please help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Alex_Xolod [135]

Answer:

Step-by-step explanation:

4y+3x

5 0
3 years ago
How I can answer this question, NO LINKS, if you answer correctly I will give u brainliest!
olga nikolaevna [1]

Answer:

27 children

Step-by-step explanation:

So, if you add 9 +4 = 13. 13 x 3 = 39.

3 times 9 = 27

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5 0
2 years ago
Read 2 more answers
35,000,000 in scientific notation
Mandarinka [93]

Answer:

3.5 times 10^7

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

3.5×10^7

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8 0
3 years ago
Read 2 more answers
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 195,
Semenov [28]
Given:
n = 195, sample size.
x = 162, successes in the sample

The proportion is
p = x/n = 162/195 = 0.8308

n* p = 195*0.8308 = 162
n*(1-p) = 195*(1 - 0.8308) = 33
If n*p >= 10, and n*(1-p) >= 10, then the sample proportions will have a normal distribution. This condition is satisfied.

The proportion mean is
μ = 0.8308
The proportion standard deviation is
\sigma =  \sqrt{ \frac{p(1-p)}{n} }  = \sqrt{ \frac{0.8308(1-0.8308)}{195} } =0.0269

σ/√n = 0.0269/√195 = 0.00192

At the 95% confidence level, the interval for the population proportion is
(μ - 1.96(σ/√n), μ + 1.96(σ/√n))
= (0.8308 - 1.96*0.00192, 0.8308 + 1.96*0.00192)
= (0.827, 0.8345)

Answer: The 95% confidence interval is (0.827, 0.835)








5 0
3 years ago
Show that p must be directed through centroids of cross sections if axial stress ? is not to vary over a cross section.
wolverine [178]
Assuming P (usually written in upper case) represents a force normal to a given cross section.

If a point load is applied to any point of the section, stress concentration will cause axial stress to vary.

The context of the question considers the uniformity of axial stress at a certain distance away from the point of application (thus stress concentration can be neglected).

If a force P is applied through the centroid, sections will be stressed uniformly.  However, if the force P is applied at a distance "e" from the centroid, the equivalent load on the section equals an axial force and a moment Pe.  The latter causes bending of the member, causing non-uniform stress.

If we assume A=(uniform) cross sectional area, and I=moment of inertia of the section, then stress varies with the distance y from the centroid equal to
stress=sigma=P/A + My/I
where P=axial force, M=moment = Pe.
Therefore when e>0, the stress varies across the section.
7 0
2 years ago
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