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Marizza181 [45]
3 years ago
8

what is the equation of a line parallel to the line 2x-3y=6 and that passes through the point (-6,12)

Mathematics
1 answer:
Brrunno [24]3 years ago
8 0

Hello from MrBillDoesMath!

Answer:

y =  (2/3)x + 16

Discussion:

Given line:

2x - 3y = 6         =>   add 3y to both sides

2x = 6 + 3y        => subtract 6 from both sides

2x -6 = 3y          => divide both sides by 3

y =(2/3)x - 2  

The slope of this line, m,  (2/3) and any line parallel to the given line has the same slope.  We are looking for the line with slope (2/3) passing through (-6, 12)

y = (2/3)x + b                                => substitute (x,y) = (-6, 12) in the equation

12 = (2/3)(-6) + b                           =>  add (2/3)(6) = 12/3 = 4

12 + 4 = (2/3)(-6) + (2/3)(6) + b     => as (2/3)(-6) + (2/3)(6)  = 0

12 +4 = 0 + b    =>

b = 16

Hence the equation of the parallel line through ( -6,12) is

y = mx + b

  =(2/3)x + 16

Check: is (-6,12) on this line?  Does 12 = (2/3)(-6) + 16 =  -4 + 16 = 12? Yes!

Thank you,

MrB

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Ahat [919]

Answer:

The soda costs 2.20 and the sandwich costs 7.70

Step-by-step explanation:

To find this, set the soda cost as x. We now know that the sandwich cost is 3.5x. Add these together and set equal to 9.90

x + 3.5x = 9.90

4.5x = 9.90

x = 2.20

This is the cost of the soda. Now we can multiply that by 3.5 to get the sandwich cost.

3.5 * 2.20 = 7.70

7 0
3 years ago
In an experiment to determine whether there is a systematic difference between the weights obtained with two different mass bala
Maslowich

Answer:

H0: μd=0 Ha: μd≠0

t= 0.07607

On the basis of this we conclude that the mean weight differs between the two balances.

Step-by-step explanation:

The null and alternative hypotheses as

H0: μd=0 Ha: μd≠0

Significance level is set at ∝= 0.05

The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Specimen        A               B           d = a - b         d²

1                     13.76        13.74         0.02           0.004

2                    12.47        12.45          0.02         0.004

3                    10.09        10.08           0.01        0.001

4                       8.91       8.92          -0.01          0.001

5                     13.57      13.54           0.03        0.009

<u>6                     12.74      12.75          -0.01        0.001</u>

<u>∑                                                      0.06         0.0173</u>

d`= ∑d/n= 0.006/6= 0.001

sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882

sd= 0.05368

t= 0.001/ 0.05368/ √6

t= 0.18629/2.449

t= 0.07607

Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.

3 0
3 years ago
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3 years ago
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c\sin 28^\circ = 27 \sin 102^\circ

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3 years ago
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