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Drupady [299]
3 years ago
14

3xc = -5, solve for x

Mathematics
1 answer:
atroni [7]3 years ago
4 0

Answer:

x = -\dfrac{5}{3c}

Step-by-step explanation:

You want x alone on the left side.

x is being multiplied by 3 and by c.

The opposite operation of multiplication is division. You must divide 3xc by 3c to end up with just x.

You must do the same operation to both sides of an equation, so divide both sides by 3c.

3xc = -5

Divide both sides by 3c.

\dfrac{3xc}{3c} = \dfrac{-5}{3c}

Simplify both sides.

x = -\dfrac{5}{3c}

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Factorization of 3x2-8x-3
pychu [463]
<span>3x</span>² <span>- 8x -3  = 3x</span>² - 9x + x - 3 =

= 3x² - 9x + x - 3 = 3x* ( x - 3) + 1* (x - 3) = 

= 3x * (x - 3) + 1 * (x - 3) = (3x + 1) (x - 3) 

Answer (3x + 1) (x - 3)

3 0
3 years ago
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I’ve been trying to move on but I can’t get through it can someone please help me out please please
nadya68 [22]

Answer:

84 - c

Step-by-step explanation:

The correct expression would be 84 - c

4 0
3 years ago
A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i
garri49 [273]

Answer:

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :  

Null hypothesis:\mu \leq 5  

Alternative hypothesis:\mu > 5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0
3 years ago
solve for each please i really need help if u want to help me with my test and i get an a or a b i will give u 500 dollars
Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

  • Every x has only one value of y
  • x appears once in ordered pairs

<u><em>Examples:</em></u>

  • The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
  • The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
  • The domain is the set of values of x
  • The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

3 0
3 years ago
Looking for some on this problem
adelina 88 [10]
The correct answer is D
8 0
3 years ago
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