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2 years ago

A set of equations is given below:

Mathematics

1 answer:

artcher [175]2 years ago

7 0

Answer:

comparing both equation we get,

2X+1=X+9

or, X=8

Y=X+9

or, Y=17

therefore the required solution is (8,17)

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Can anybody please help me ?

Ivenika [448]

Answer:

R=5

H=12

D=10

Step-by-step explanation:

4 0

2 years ago

If AD = 6 cm, DB = 4.2 cm, AB = 8.7 cm, and ED = 3.9 cm, what is the measure of DC?

marshall27 [118]

You cannot work this out without the angles

4 0

3 years ago

Simplify |-5| + |6|. A) -30 B) -1 C) 1 D) 11

evablogger [386]

The Answer is D)11! ;)

4 0

2 years ago

Please help I will mark brainlist if correct

Virty [35]

We have 2 equations here ( b = cost of bush and t = cost of a tree)

10b + 4t = 246........................................................(1)
5b + 3t = 147 Multiply this equation by -2:-
-10b - 6t = -294.......................................................(2)

Adding equations (1) and (2) we get:-

0 - 2t = - 48

t = -48/-2 = 24

Plug t = 24 into the first equation:-
10b + 4(24) = 246
10b = 246 - 96 = 150
b = 15

The answer is one bush costs $15 and one tree costs $24.

4 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