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2 years ago

Which ordered pair is a solution of the equation -3+5y=2x+3y ?

Mathematics

1 answer:

Oksana_A [137]2 years ago

8 0

Answer: x = y + −3/2

Step-by-step explanation:

Let's solve for x.

−3+5y=2x+3y

Step 1: Flip the equation.

2x+3y=5y−3

Step 2: Add -3y to both sides.

2x+3y+−3y=5y−3+−3y

2x=2y−3

Step 3: Divide both sides by 2.

2x/2 = 2y - 3/2

ANSWER: x = y + -3/2

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For a process, the average range for all samples was 5 and the process average was 25. If the sample size was 10, calculate UCL

alexira [117]

Answer:

8.885

Step-by-step explanation:

Given that :

Sample size, n = 10

The average range, Rbar for all samples = 5

The upper control limit, UCL for the R-chart is :

UCL L= D4Rbar

From the control chart constant table, D4 = 1.777

Hence,

UCL = 1.777 * 5

UCL = 8.885

The UCL for the R-chart is 8.885

7 0

3 years ago

What is the solution for x>16

Volgvan

Answer:

The answer would be (16 , ∞)

Step-by-step explanation:

7 0

3 years ago

Which is the range of the function f(x) = (9)×?

Anna [14]

Using it's concept, it is found that the range of the function $f(x) = 9^x$ is given by:

all real numbers greater than 0.

<h3>What is the range of a function?</h3>

It is the set that contains all possible output values for the function.

In this problem, the function is:

$f(x) = 9^x$

For <u>exponential functions without vertical shifts</u>, as is the case in this problem, the range is given by:

all real numbers greater than 0.

More can be learned about the range of a function at brainly.com/question/10891721

#SPJ1

4 0

2 years ago

1.Which Symbol will make the number sentence - | 16 | _ 16 true?

Dahasolnce [82]

1. =
2.
3.q+4=-4
I don't know number 2

4 0

3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago