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1 year ago

Which graph represents the solution8 - 3x ≥ -4

Mathematics

1 answer:

Vedmedyk [2.9K]1 year ago

6 0

Answer:

3rd option

Step-by-step explanation:

8 - 3x ≥ - 4 ( subtract 8 from both sides )

- 3x ≥ - 12

divide both sides by - 3 , reversing the symbol as a result of dividing by a negative quantity.

x ≤ 4

since ≤ the circle on number line is solid ( not open )

since less than 4 then arrow points left away from 4

the solution is shown on the third graph

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Can someone please help

iren [92.7K]

So the sides ML and MN are exactly the same. Your equation would look something like this...
3x = x + 10
You first subtract x from both sides to isolate the variable.
2x = 10
Divide 2 from both sides.
x = 5
5 is the length of MN.
I hope this helps love! :)

3 0

2 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured parallel to a line

Ber [7]

Answer:

The distance is:

$\displaystyle\frac{3\sqrt{142}}{10}$

Step-by-step explanation:

We re-write the equation of the line in the format:

$\displaystyle\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}$

Notice we divided the fraction of y by 2/2, and the fraction of z by 3/3.

In that equation, the director vector of the line is built with the denominators of the equation of the line, thus:

$\displaystyle\vec{v}=\left< 3, 2, \frac{5}{3}\right>$

Then the parametric equations of the line along that vector and passing through the point (-2, 3, -4) are:

$x=-2+3t\\y=3+2t\\\displaystyle z=-4+\frac{5}{3}t$

We plug them into the equation of the plane to get the intersection of that line and the plane, since that intersection is the image on the plane of the point (-2, 3, -4) parallel to the given line:

$\displaystyle x+y+z=3\to -2+3t+3+2t-4+\frac{5}{3}t=3$

Then we solve that equation for t, to get:

$\displaystyle \frac{20}{3}t-3=3\to t=\frac{9}{10}$

Then plugging that value of t into the parametric equations of the line we get the coordinates of the intersection:

$\displaystyle x=-2+3\left(\frac{9}{10}\right)=\frac{7}{10}\\\displaystyle y=3+2\left(\frac{9}{10}\right)=\frac{24}{5} \\\displaystyle z=-4+\frac{5}{3}\left(\frac{9}{10}\right)=-\frac{5}{2}$

Then to find the distance we just use the distance formula:

$\displaystyle d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

So we get:

$\displaystyle d=\sqrt{\left(-2-\frac{7}{10}\right)^2+\left(3-\frac{24}{5}\right)^2+\left(-4 +\frac{5}{2}\right)^2}=\frac{3\sqrt{142}}{10}$

3 0

3 years ago

What number am I?

Nataliya [291]

Answer:

1.11

Step-by-step explanation:

I believe its 1.11

3 0

3 years ago

Read 2 more answers

What is the ratio of the intensities of an earthquake PP wave passing through the Earth and detected at two points 19 kmkm and 4

Tamiku [17]

Answer:

The ratio of the intensities is roughly 6:1.

Step-by-step explanation:

The intensity I() of an earthquake wave is given by:

$I = \frac{P}{4\pi d^{2}}$

<em>where P: is the power ans d: is the distance. </em>

Hence, the ratio of the intensities of an earthquake wave passing through the Earth and detected at two points 19 km and 46 km from the source is:

$\frac{I_{1}}{I_{2}} = \frac{P/4\pi d_{1}^{2}}{P/4\pi d_{2}^{2}}$