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3 years ago

What is the order from least to greatest of the expression x, x-2, x2, 0?

Mathematics

1 answer:

netineya [11]3 years ago

8 0

Answer:

i think the answer would be x-2,0,x,x2

Step-by-step explanation:

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Would be very much appreciated if you helped out :)

Tanya [424]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Can someone help with this math problem? 3 (x+6) = 27

Ainat [17]

Answer:

x = 3

Step-by-step explanation:

Use distributive property: a(b +c) =(a*b) + (a*c)

3(x + 6) = 27

3*x + 3*6 = 27

3x + 18 = 27 {Subtract 18 from both sides}

3x + 18 -18 = 27 - 18

3x = 9 {Divide both sides by 3}

3x/3 = 9/3

x = 3

7 0

2 years ago

What is the expression that means the same as 7 less than the square of a number? A. 7-2x B. 7-2× C. 7-x2 D. x2-7

damaskus [11]

Answer d is correct!!!!!

4 0

3 years ago

Read 2 more answers

A random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph. From past information, it is kn

loris [4]

Answer:

95 percent confidence interval for μ is determined by

(60.688 , 69.312)

a) be the same

Step-by-step explanation:

Step (i):-

Given data a random sample of 121 automobiles traveling on an interstate showed an average speed of 65 mph.

The sample size is n= 121

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (ii) :-

Given The 95 percent confidence interval for μ is determined as

(61.08, 68.92)

we are to reduce the sample size to 100 (other factors remain unchanged) so

given The sample size is n= 100

The mean of the sample x⁻ = 65mph.

The standard deviation of the population is 22 mph

σ = 22mph

Step (iii):-

95 percent confidence interval for μ is determined by

$(x^{-} - 1.96\frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} } )$

$(65 - 1.96\frac{22}{\sqrt{100} } , 65 + 1.96 \frac{22}{\sqrt{100} } )$

(65 - 4.312 ,65 +4.312)

(60.688 , 69.312) ≅(61 ,69)

This is similar to (61.08, 68.92) ≅(61 ,69)

Conclusion:-

it become same as (61.08, 68.92)

4 0

3 years ago

Lim ln(tan x) as tends to pi/2 from the left

ioda

Let $x=\arctan y$ . Then $\tan x=y$ , and so as $x\to\dfrac\pi2^-$ , you have $y\to+\infty$ . The limit is then equivalent to

$\displaystyle\lim_{x\to\frac\pi2^-}\ln(\tan x)=\lim_{y\to\infty}\ln y=\infty$

3 0

3 years ago