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artcher [175]
3 years ago
6

My number had a tens digit that is 8 more than the ones digit. Zero is not one of my digits

Mathematics
1 answer:
marin [14]3 years ago
4 0

The tens digit is 8 more than the ones digit. There isn't any zeros.

The maximum digit is 9 because 10 would be 2 digits, 1 and 0.

So say the tens digits is 9.

9 _

The ones digit is 8 less so 9-8 = 1

9 1

There isn't any zeros which is correct.

The answer is 91.  

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Can the law of sines be used to solve the triangle
ivanzaharov [21]

Answer:

No

Step-by-step explanation:

No, the law of sines cannot be used to solve the triangle. The triangle shows the measures of two sides and an included angle. To use the law of sines, you need to know the measure of an angle and its opposite side.

Answer On Edg. 2020

8 0
3 years ago
A system has two failure modes. One failure mode, due to external conditions, has a constant failure rate of 0.07 failures per y
nadya68 [22]

Answer:

0.9177

Step-by-step explanation:

let us first represent the two failure modes with respect to time as follows

R₁(t) for external conditions

R₂(t) for wear out condition ( Wiebull )

Now,

R1(t) = e^{-nt} .....1

where t = time in years = 1,

n = failure rate constant = 0.07

Also,

R2(t)=e^{-(\frac{t}{Q} )^{B} }......2

where t = time in years = 1

where Q = characteristic life in years = 10

and B = the shape parameter = 1.8

Substituting values into equation 1

R1(t) = e^{-(0.07)(1)} \\\\R1(t) = e^{-0.07}

Substituting values into equation 2

R2(t)=e^{-(\frac{1}{10} )^{1.8} }\\\\R2(t)=e^{-(0.1)}^{1.8} }\\\\R2(t)=e^{-0.0158}

let the <em>system reliability </em>for a design life of one year be Rs(t)

hence,

Rs(t) = R1(t) * R2(t)

t = 1

Rs(1) = [e^{-0.07} ] * [e^{-0.0158} ] = 0.917713

Rs(1) = 0.9177 (approx to four decimal places)

5 0
3 years ago
What is the equation for 5 more than 3 times a number is 0
Sedaia [141]

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Step-by-step explanation:

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sergejj [24]

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3 years ago
What is the solution to the inequality 2n+5| &gt; 1?
hodyreva [135]

Answer:

n < - 3 or n > - 2

Step-by-step explanation:

Inequalities of the type | x | > a , have solutions of the form

x < - a or x > a

Then

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Solve both inequalities

2n + 5 < - 1 ( subtract 5 from both sides )

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n < - 3

OR

2n + 5 > 1 ( subtract 5 from both sides )

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Solution is n < - 3 or n > - 2

3 0
3 years ago
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