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3 years ago

I need equivalent expression to this x+x+x+2

1 answer:

5 0

The X, is added together 3 times but can just be multiplied by 3 since it’s the same variable. The 2 must remain Uncombined because it cannot join the unknown variable (x). Answer is 3x+2

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6789 divided by 89.65

Sergeeva-Olga [200]

Step-by-step explanation:

75 . 727 is your answer

hope it is helpful to you

5 0

3 years ago

Read 2 more answers

For a population with a mean of u = 40 and a standarddeviation of o = 8, find the z-score corresponding toeach of the following

ivann1987 [24]

Solution:

Given the mean and standard deviation;

$\mu=40,\sigma=8$

The z-score formula is;

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

(a)

$\begin{gathered} x=34,n=1 \\ \\ z=\frac{34-40}{\frac{8}{\sqrt{1}}} \\ \\ z=-0.75 \end{gathered}$

(b)

$\begin{gathered} x=34,n=4 \\ \\ z=\frac{34-40}{\frac{8}{\sqrt{4}}} \\ \\ z=-1.5 \end{gathered}$

(c)

$\begin{gathered} x=34,n=16 \\ \\ z=\frac{34-40}{\frac{8}{\sqrt{16}}} \\ \\ z=-3 \end{gathered}$

5 0

1 year ago

Find a such that the solution of y'' + y' − 2y = 0, y(0) = a, y' (0) = 1 tends to zero as t → +∞.

Kay [80]

Answer:

Step-by-step explanation:

$y'' + y' − 2y = 0, y(0) = a, y' (0) = 1$

Auxialary equation is

$m^2+m-2=0\\m=-2,1$

General solution is

$y=Ae^{-2x} +Be^x$

$y(0) = A+B =a$

$y'(0) = -2Ae^{2x} +Be^x = -2A+B = 1$

Eliminate B to get

3A =a-1

We know that y tends to 0 when x tends to infinity for any finite A

i.e. a should be a finite real number.

8 0

3 years ago

20 POINTS: solve for x.

lara [203]

Answer:x=7

Step-by-step explanation:

Here AD=6

AB=6 (tangents drawn from external point to a circle are equal)

Then BC=13-6=7

Now, CE=BC=7(tangents drawn from external point to a circle are equal)

EF=14-7=7

EF=FG=x=7(tangents drawn from external point to a circle are equal)

3 0

3 years ago

A line passes through the point (–6, –3) and has a slope of 2/3 . Which point is on the same line?

navik [9.2K]

Answer:

The point (0, 1) represents the y-intercept.

Hence, the y-intercept (0, 1) is on the same line.

Step-by-step explanation:

We know that the slope-intercept form of the line equation

y = mx+b

where

m is the slope

b is the y-intercept

Given

The point (-6, -3)

The slope m = 2/3

Using the point-slope form

$y-y_1=m\left(x-x_1\right)$

where

m is the slope of the line

(x₁, y₁) is the point

In our case:

m = 2/3

(x₁, y₁) = (-6, -3)

substituting the values m = 2/3 and the point (-6, -3) in the point-slope form

$y-y_1=m\left(x-x_1\right)$

$y-\left(-3\right)=\frac{2}{3}\left(x-\left(-6\right)\right)$

$y+3=\frac{2}{3}\left(x+6\right)$

Subtract 3 from both sides

$y+3-3=\frac{2}{3}\left(x+6\right)-3$

$y=\frac{2}{3}x+4-3$

$y=\frac{2}{3}x+1$

comparing with the slope-intercept form y=mx+b

Here the slope = m = 2/3

Y-intercept b = 1

We know that the value of y-intercept can be determined by setting x = 0, and determining the corresponding value of y.

Given the line

$y=\frac{2}{3}x+1$

at x = 0, y = 1

Thus, the point (0, 1) represents the y-intercept.

Hence, the y-intercept (0, 1) is on the same line.

5 0

3 years ago