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4 years ago

For which values of x is the inequality 3(1 + x) > x + 17 true?

Mathematics

1 answer:

Snezhnost [94]4 years ago

7 0

a. 3(1 + 2) = 9; 2 + 17 = 19

9 > 19 is false so A is wrong, regardless if 3 works.

b. 3(1 + 4) = 15; 4 + 17 = 21

15 > 21 is false so we know B is wrong, regardless if 5 works.

c. 3(1 + 6) = 21; 6 + 17 = 23

21 > 23 doesn't work either.

d. 3(1 + 8) = 27; 8 + 17 = 25

27 > 25 is true. So let's test out 9.

3(1 + 9) = 30, 9 + 17 = 26

30 > 26 is also true so D is correct.

Distribute: 5x-15-6x+12 > 8x +2

Simplify: -x-3 > 8x + 2

Make one side 0: 9x + 5 < 0

From there you know your answer cannot be positive or it will be above zero so A, B, and C are wrong because they all allow positive numbers at some point. So D is correct. (again)

I'm not entirely sure about this one so this is my thought.

Distribute: 2 -6x - 3 < 12x -24x

Simplify: -6x -1 < -12x

Set one side to 0: 6x - 1 < 0

From there, you know the answer isn't A, B, or C because plugging in 1/6 doesn't work. So I will say that D is right as every number less than -1/6 will work.

Fahrenheit Question:

You can say that A is wrong because the equal to means that the person may be at normal temperature and not hyperthermia. That also makes B wrong as well because it too uses the equal to. C uses the less than sign (in comparison to 98.6) meaning that it looks for Celsius temperatures below the normal F temp which isn't hyperthermia. D does use the greater sign signaling that it is a temperature above the normal 98.6 So I say the answer is (again and a little fearfully) D.

Last Question:

You know their is no solution because both variables are equal but the constants are different. since -5r is equal to -5r they will ultimately cancel each other out and you'll be left with 6 is less than or equal to -10 which can never be true if the variables are the same. That means the answer is (I say very fearfully now) D.

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Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$