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Bumek [7]
3 years ago
15

Can you help me? I will give brainliest!

Mathematics
1 answer:
zzz [600]3 years ago
6 0

Answer:

√1

Step-by-step explanation:

If you do the fractions by squaring them then you get fractions that equal 1:

5/13^2 = 25/169

12/13^2 = 144/169

144/169 + 25/169 = 169/169 = 1

√1

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Unless you meant -1/2x+2 the top answer is correct. If you meant the other it is -1/2(d/dx(x+2), which results in -1/2(1) or -1/2.
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What is the least 6 digit multiple of 4 that has all different numbers?
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Answer:

We can now take 2 as the third digit, 3 as the fourth digit, 4 as the fifth digit and finally 5 as the sixth digit. Therefore, the smallest 6-digit number having all different digits is 102345.

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Are the fraction 5/6,5/7,5/8,and 5/9 arranged in order from least to greatest or from greatest to least ?
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4 years ago
Prove that the points (a, -3a), (2a, a) and (0, -2a) form a scalene trqingle​
Dmitry [639]

Answer:

See below.

Step-by-step explanation:

Remember that a scalene triangle has lengths of different values.

Therefore, we just need to find the length or <em>distance</em> from each point to the next. If the three distances we acquire are different, then we prove that the point do indeed form a scalene triangle.

Let's let A be (a, -3a), B be (2a, a), and C be (0, -2a).

So, let's find each of the side lengths using the distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Side AB:

Let's let A:(a, -3a) be (x₁, y₁) and let's let B:(2a, a) be (x₂, y₂). Substitute this into our formula:

d=\sqrt{(2a-a)^2+(a-(-3a))^2

Subtract:

d=\sqrt{(a)^2+(4a)^2

Square:

d=\sqrt{a^2+16a^2}

Add:

d=\sqrt{17a^2}

Simplify:

d=\sqrt{a^2}\cdot \sqrt{17}\\d=|a|\sqrt{17}

So:

\overline {AB}=|a|\sqrt{17}

Note: We need the absolute value because anything squared will be positive, and if you take the square root of something positive, the result will be positive. The absolute value ensures that the a value will be positive no matter what a is to begin with.

Side BC:

Let's let C:(0, -2a) be (x₁, y₁) and let's let B:(2a, a) be (x₂, y₂).

d=\sqrt{(2a-0)^2+(a-(-2a)^2}

Subtract:

d=\sqrt{(2a)^2+(3a)^2}

Square:

d=\sqrt{4a^2+9a^2}

Add:

d=\sqrt{13a^2}

Simplify:

d=\sqrt{a^2}\cdot \sqrt{13}\\d=|a|\sqrt{13}

Therefore:

\overline{BC}=|a|\sqrt{13}

Side AC:

Let's let A:(a, -3a) be (x₁, y₁) and let's let C:(0, -2a) be (x₂, y₂).

d=\sqrt{(0-a)^2+(-2a-(-3a))^2

Subtract:

d=\sqrt{(-a)^2+(a)^2}

Square:

d=\sqrt{a^2+a^2}

Add:

d=\sqrt{2a^2}

Simplify:

d=|a|\sqrt2

Therefore:

\overline{AC}=|a|\sqrt2

So, our three side lengths are:

\overline {AB}=|a|\sqrt{17}\text{, }\overline{BC}=|a|\sqrt{13}\text{, and } \overline{AC}=|a|\sqrt2

We can see that the three side lengths are different since they do not equal to same thing.

Therefore, we can deduce that the triangle must be scalene.

And we're done!

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Inverse both sides

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♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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