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Oliga [24]
3 years ago
13

Fiona wrote the linear equation y = 2/5 X -5. When Henry wrote his equation they discovered that his equation had all the same S

olutions as fionas. which equation could have been Henry's.
A. X- 5/4y =25/4
B. X-5/2y=25/4
C. X-5/4y =25/4
D. X- 5/2y=25/2

Mathematics
2 answers:
REY [17]3 years ago
5 0
The equation of the graph is given as y = (2/5)x - 5.

You have to figure out which of the choices equals the equation given,
y = (2/5)x - 5.
You could solve each of the answer choices for y, but since each choice is in the format x - (m)y = b, you can put the given equation in that format too.

y= \frac{2}{5}x - 5 \\ 
5= \frac{2}{5}x-y \\ 
 \frac{2}{5}x - y = 5 \\ 
 (\frac{5}{2}) \frac{2}{5}x - (\frac{5}{2})y = (\frac{5}{2})5 \\ 
x -  (\frac{5}{2})y =  \frac{25}{2}

katrin2010 [14]3 years ago
3 0

Answer: D

Step-by-step explanation:

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Answer:

The answer is "(\frac{\pi}{2})".

Step-by-step explanation:

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Let add equation 1 and 2:

Using formula: x^2+y^2=1 \\\\

convert to polar coordinates

r=2

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    =\int^{2\pi}_{0}\int^{1}_{0}  (Z_2-z_1)r  \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r)  \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0}  \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4})  \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\

8 0
3 years ago
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4 0
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saul85 [17]
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d1i1m1o1n [39]

Answer:

The equations shows a difference of squares are:

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Step-by-step explanation:

the difference of two squares is a squared number subtracted from another squared number, it has the general from Ax² - By²

We will check the options to find which shows a difference of squares.

1) 10y²- 4x²

The expression is similar to the general form, so the equation represents a difference of squares.

It can be factored as (√10 y + 2x )( √10 y - 2x)

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The expression is similar to the general form, so the equation represents a difference of squares.

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The expression is not similar to the general form, so the equation does not represent a difference of squares.

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