Ask question

Ask question

All categories

4 years ago

15

Select the correct answer. What are the values of x in this equation? `x^(2/5) + x^(1/5) + 1 = 3` A. 32, -1 B. 32, 1 C. -32, 1 D

. -32, -1

1 answer:

mrs_skeptik [129]4 years ago

4 0

x^(2/5) + x^(1/5) + 1 = 3

subtract 3 from each side

x^(2/5) + x^(1/5) -2=0

let m = x^1/5

m^2 +m -2 =0

(m - 1) (m + 2) = 0

using the zero product property

m=1 m=-2

replace m = x^1/5

x^1/5 =1 and x^1/5 = -2

take everything to the 5th powr

x^ 1/5 ^ 5 = 1^5 and x^ 1/5 ^ 5 = -2 ^ 5

x = 1 and x = -32

Choice C

C. -32, 1

You might be interested in

What is the remainder when 3 is synthetically divided into the polynomial -x^2+5x-9

Maurinko [17]

-3 is the remainder when you synthetically divide

8 0

3 years ago

Read 2 more answers

How to solve a variable?

amm1812

Answer:

If the equation is in the form, ax + b = c, where x is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division. Solve 3y + 2 = 11. Subtract 2 from both sides of the equation to get the term with the variable by itself.

Hope this helps :D

7 0

3 years ago

For all values of x

Yuliya22 [10]

Answer:

A.) gf(x) = 3x^2 + 12x + 9

B.) g'(x) = 2

Step-by-step explanation:

A.) The two given functions are:

f(x) = (x + 2)^2 and g(x) = 3(x - 1)

Open the bracket of the two functions

f(x) = (x + 2)^2

f(x) = x^2 + 2x + 2x + 4

f(x) = x^2 + 4x + 4

and

g(x) = 3(x - 1)

g(x) = 3x - 3

To find gf(x), substitute f(x) for x in g(x)

gf(x) = 3( x^2 + 4x + 4 ) - 3

gf(x) = 3x^2 + 12x + 12 - 3

gf(x) = 3x^2 + 12x + 9

Where

a = 3, b = 12, c = 9

B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)

To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula

Y = 3x + 3

X = 3y + 3

Make y the subject of formula

3y = x - 3

Y = x/3 - 3/3

Y = x/3 - 1

Therefore, g'(x) = x/3 - 1

For g'(12), substitute 12 for x in g' (x)

g'(x) = 12/4 - 1

g'(x) = 3 - 1

g'(x) = 2.

5 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

C

b

C

c

C

d

A

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

Determine if the sequence is geometric. If it is, find the common ratio.

Akimi4 [234]

Answer:

-1/3
not geometric
-5
4

Step-by-step explanation:

You find the ratio of terms by dividing the second by the first. If the ratio is the same for all adjacent pairs of terms, it is a "common" ratio.

1) (2/3)/-2 = -1/3

(-2/9)/(2/3) = -1/3

(2/27)/(-2/9) = -1/3 . . . . . the common ratio is -1/3

___

2) 2/0 = undefined . . . . not a geometric sequence

___

3) -10/2 = -5

50/-10 = -5

-250/50 = -5 . . . . . the common ratio is -5

___

4) 4/1 = 4

16/4 = 4

64/16 = 4 . . . . . the common ratio is 4

8 0

3 years ago