Find the equation of a straight line that is parallel to 4y-3x+10=0

1 answer:

5 0

First, put the equation in slope intercept form: y=mx+b

4y-3x+10=0
Subtract 4y from both sides: (the zero cancels out because it's nothing)

-3x +10=-4y
Divide both sides by -4 to get y alone:

3/4x+5/2=y or y=3/4x+5/2 (they are the same thing flipped either way).

Now you have y=3/4x+5/2

To find a parallel line, you have to know that ALL parallel lines will have the same slope as the first line.

so the slope will be 3/4x

For the y intercept, you can choose any number you want OTHER than the one that was on the previous line. For example, you can have any of these equations to make it parallel to the first line:

y=3/4x+3
y=3/4x-6.55552
y=3/4x+10000259
y=3/4x-67487987.009999

Literally, the y intercept could be anything but 5/2

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The Sahara Desert has an area of approximately 9,400,400 km2. While estimates of its average depth vary, they center around 150

dolphi86 [110]

Answer:

$1.128048 \times 10^{25}$ grains of sand are in the Sahara Desert

Step-by-step explanation:

Given :

The Sahara Desert has an area of approximately 9,400,400 sq.km.

Depth = 150 m

To Find :Approximately how many grains of sand are in the Sahara Desert?

Solution:

Area of Sahara desert = 9,400,400 sq.km.

$1 km^2 = 1000000 m^2$

So, $9,400,400 km^2 = 1000000 \times 9400400 m^2$

$9,400,400 km^2 =9.4004 \times 10^{12}m^2$

Volume of sand = $Area \times depth$

= $1.41006\times 10^{15} m^3$

$1 m ^3 = 1000000 cm^3$

So, $1.41006\times 10^{15} m^3 = 1.41006\times 10^{15} \times 1000000cm^3$

$1.41006\times 10^{15} m^3 =1.41006 \times 10^{21}cm^3$

1 cubic cm holds grains of sand = 8000

$1.41006 \times 10^{21}cm^3$ holds grains of sand = $1.41006 \times 10^{21} \times 8000$

= $1.128048 \times 10^{25}$

Hence $1.128048 \times 10^{25}$ grains of sand are in the Sahara Desert

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42 divided by 4 and 67 divided by8

Bezzdna [24]

10.5 and 8.4 ......................,,,,,

6 0

3 years ago

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There is a photo attached tan 270 is incorrect!

wlad13 [49]

$\dfrac{1-\cos135^\circ}{\sin135^\circ}=\dfrac{1-\left(\cos^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{1-\left(1-\sin^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\medskip\\=\dfrac{2\sin^2 67.5^\circ}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{\sin 67.5^\circ}{\cos 67.5^\circ}=\tan 67.5^\circ$