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2 years ago

8

If f(x) = 3X + 10 and g(x) = 2x – 4, find (f- g)(x).

1 answer:

svetoff [14.1K]2 years ago

7 0

So (f-g)(x) is f(x)-g(x) so you have

(3x+10)-(2x-4)

Distribute the negative in to the 2x-4 to get

3x+10-2x+4

Then combing like terms you have

3x-2x+10+4

And then that gives you

x+14

So (f-g)(x)=x+14

Hope this helps!! Mark as brainliest!!!!

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Solve the system by elimination.(show your work)

PilotLPTM [1.2K]

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation by elimination:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

6 0

3 years ago

Read 2 more answers

The article "Occurrence and Distribution of Ammonium in Iowa Groundwater" (K. Schilling, Water Environment Research, 2002:177–18

Papessa [141]

Answer: 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

Step-by-step explanation:

Since we have given that

Number of alluvial wells = 349

Number of quaternary wells = 143

Number of alluvial wells that had concentrations above 0.1 = 182

Number of quaternary wells that had concentrations above 0.1 = 112

Average of alluvial wells = 0.27

Standard deviation = 0.4

Average of quaternary wells = 1.62

Standard deviation =1.70

So, 95% confidence interval gives

alpha = 5% level of significance.

$\dfrac{\alpha}{2}=2.5\%\\\\z_{\frac{\alpha}{2}}=1.96$

So, 95% confidence interval becomes,

$(1.62-0.27)\pm 1.96\sqrt{\dfrac{0.4^2}{349}+\dfrac{1.7^2}{143}}\\\\=1.35\pm 1.96\times 0.020\\\\=(1.35-0.040,1.35+0.040)\\\\=(1.31,1.39)$

Hence, 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

6 0

3 years ago

Which of the following values are in the range of the function graphed below?

Dafna1 [17]

Answer:

A. 1

Step-by-step explanation:

I see only one red function line in the graph. this line represents the simple function y = 1, for x in the interval [-2,1].

the domain of a function defines the possible values for x, and the range of a function defined the possible values for y.

this function has only one possible value for y : 1

so, only A applies.

5 0

3 years ago

Write an equation in

Nastasia [14]

Answer:

y+6=-3/4(x-2)

Step-by-step explanation:

y-(-6)=-3/4(x-2)

y+6=-3/4(x-2)

7 0

2 years ago

i need help!!! Match each power of ten with the appropriate number. 1. 100 2. 102 3. 10-3 4. 10-2 a. 0.01 b. 0.001 c. 1 d. 100

Masja [62]

1. 100 -> 10²

2. 102 -> 10² + 2

3. 10^-3 -> 1/(1000)

4. 10^-2 -> 1/(100)

a. 0.01 -> 10^-2

b. 0.001 -> 10^-3

c. 1 -> 1

d. 100 -> 10^2

hope this helps

5 0

3 years ago