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Tanzania [10]
3 years ago
14

Classify the function as linear or quadratic and identify the quadratic linear and constant terms. PLEASE HELP ME!!!

Mathematics
1 answer:
patriot [66]3 years ago
5 0
The equation
y =  - 3 {x}^{2}  - 29x + 30
is quadratic, since the highest power of x is x^2. The quadratic term is -3x^2, the linear term is -29x, and the constant is 30.
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Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.
xz_007 [3.2K]

Answer:

Part 1) f(x)=\frac{2x-1}{x+2} -------> f^{-1}(x)=\frac{-2x-1}{x-2}

Part 2) f(x)=\frac{x-1}{2x+1} -------> f^{-1}(x)=\frac{-x-1}{2x-1}

Part 3) f(x)=\frac{2x+1}{2x-1} -----> f^{-1}(x)=\frac{x+1}{2(x-1)}

Part 4) f(x)=\frac{x+2}{-2x+1} ----> f^{-1}(x)=\frac{x-2}{2x+1}

Part 5) f(x)=\frac{x+2}{x-1} -------> f^{-1}(x)=\frac{x+2}{x-1}

Step-by-step explanation:

Part 1) we have

f(x)=\frac{2x-1}{x+2}

Find the inverse  

Let

y=f(x)

y=\frac{2x-1}{x+2}

Exchange the variables x for y and t for x

x=\frac{2y-1}{y+2}

Isolate the variable y

x=\frac{2y-1}{y+2}\\ \\ xy+2x=2y-1\\ \\xy-2y=-2x-1\\ \\y[x-2]=-2x-1\\ \\y=\frac{-2x-1}{x-2}

Let

f^{-1}(x)=y

f^{-1}(x)=\frac{-2x-1}{x-2}

Part 2) we have

f(x)=\frac{x-1}{2x+1}

Find the inverse  

Let

y=f(x)

y=\frac{x-1}{2x+1}

Exchange the variables x for y and t for x

x=\frac{y-1}{2y+1}

Isolate the variable y

x=\frac{y-1}{2y+1}\\ \\2xy+x=y-1\\ \\2xy-y=-x-1\\ \\y[2x-1]=-x-1\\ \\y=\frac{-x-1}{2x-1}

Let

f^{-1}(x)=y

f^{-1}(x)=\frac{-x-1}{2x-1}

Part 3) we have

f(x)=\frac{2x+1}{2x-1}

Find the inverse  

Let

y=f(x)

y=\frac{2x+1}{2x-1}

Exchange the variables x for y and t for x

x=\frac{2y+1}{2y-1}

Isolate the variable y

x=\frac{2y+1}{2y-1}\\ \\2xy-x=2y+1\\ \\2xy-2y=x+1\\ \\y[2x-2]=x+1\\ \\y=\frac{x+1}{2(x-1)}

Let

f^{-1}(x)=y

f^{-1}(x)=\frac{x+1}{2(x-1)}

Part 4) we have

f(x)=\frac{x+2}{-2x+1}

Find the inverse  

Let

y=f(x)

y=\frac{x+2}{-2x+1}

Exchange the variables x for y and t for x

x=\frac{y+2}{-2y+1}

Isolate the variable y

x=\frac{y+2}{-2y+1}\\ \\-2xy+x=y+2\\ \\-2xy-y=-x+2\\ \\y[-2x-1]=-x+2\\ \\y=\frac{-x+2}{-2x-1} \\ \\y=\frac{x-2}{2x+1}

Let

f^{-1}(x)=y

f^{-1}(x)=\frac{x-2}{2x+1}

Part 5) we have

f(x)=\frac{x+2}{x-1}

Find the inverse  

Let

y=f(x)

y=\frac{x+2}{x-1}

Exchange the variables x for y and t for x

x=\frac{y+2}{y-1}

Isolate the variable y

x=\frac{y+2}{y-1}\\ \\xy-x=y+2\\ \\xy-y=x+2\\ \\y[x-1]=x+2\\ \\y=\frac{x+2}{x-1}

Let

f^{-1}(x)=y

f^{-1}(x)=\frac{x+2}{x-1}

7 0
3 years ago
Can someone help me with 3 and 4 will mark brainiest answer
insens350 [35]
Number three is the first option
8 0
3 years ago
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We want to know if there is a difference between the mean list price of a three bedroom home, , and the mean list price of a fou
Vitek1552 [10]

Answer:

The alternative hypothesis will state that there is significant difference between the mean list price of a three bedroom home and the mean list price of a four bedroom home.

H_a:\mu_1-\mu_2\neq 0

Step-by-step explanation:

This would be an two-sample hypothesis test for the difference between two means.

As we are looking for differences, we are not testing if one population mean is bigger than the other. This will  be a two-tailed test and the alternative hypothesis will have a unequal sign.

The alternative hypothesis will state that there is significant difference between the mean list price of a three bedroom home and the mean list price of a four bedroom home.

This can be written as:

H_a:\mu_1-\mu_2\neq 0

meaning that the population means are significantly different.

3 0
4 years ago
Least common factor how to do in 121,99​
Blizzard [7]

Answer:

Step-by-step explanation:

Prime factorize 121 and 99

121 = 11 * 11

99 = 11 * 3 * 3

Common factor = 11

6 0
3 years ago
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All proofs start as.
d1i1m1o1n [39]

Answer:

I think its B counterexample

Step-by-step explanation:

8 0
3 years ago
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