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swat32
3 years ago
13

Please help if x = 3 and y=-4 find x2 + y2 and x2 -y2

Mathematics
1 answer:
lions [1.4K]3 years ago
6 0
1. Plug in the 3 in all the x’s that are in the equation:

(3)2 + y2 = for x2 + y2

(3)2 - y2 = for x2 - y2

This would get you for both equations..

6 + y2 = for x2 + y2

6 - y2 = for x2 - y2

2. Now let’s plug in the value of y which is -4:

6 + (-4)2 = for x2 + y2

6 - (-4)2 = for x2 + y2

This would get you for both equations...

-2 = x2 + y2

14 = x2 - y2


Hope this helped, have an awesome day!





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The length of a rectangle is four times its width. If the area of the rectangle is 324m^2, find its perimeter.
Anuta_ua [19.1K]

Answer:

perimeter=90

Step-by-step explanation:

We know that the length is four times the width, so:

l=4w

We also know the area, which is 324 m². The formula for area:

A=l*w

Insert the known values:

324=(4w)*w

Solve for w. Simplify by removing parentheses:

324=4w*w\\324=4w^2

Divide 4 from both sides to isolate the variable:

\frac{324}{4}=\frac{4w^2}{4}  \\\\81=w^2

Find the square root of both sides:

\sqrt{81} =\sqrt{w^2} \\\\w=9

The width is 9 m.

We know the width. Now find the length by using the area formula and inserting known values:

324=l*9

Solve for l. Divide both sides by 9:

\frac{324}{9}=\frac{l*9}{9}\\\\  l=36

The length of the rectangle is 36. (You can check: 4 times 9 is 36)

Now find the perimeter:

P=2l+2w

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P=2(36)+2(9)\\\\P=72+18\\\\P=90

The perimeter is 90 m.

5 0
3 years ago
Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (
Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

Spam Email be S

And, test spam positive be T

Given that

P(S) = 0.3

P(\frac{T}{S}) = 0.95

P(\frac{T}{S^c}) = 0.05

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

= 0.3

P(S^c) =  1 - P(S)

= 1 - 0.3

= 0.7

ii. P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}

= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }

= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}

= 0.8906

iii. P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}

= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }

= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}

= 0.0221

We simply applied the above formulas so that the each part could come

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3 years ago
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0.81 x 10^4
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The possiblities that exist:

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H, T, H   <----
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The answer is 1/8. Now the problem that you gave us isn't dependent upon anything, so more specifically the answer is 1/8 independent.

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Step-by-step explanation:

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