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4 years ago

I need the answer to number three please

Mathematics

1 answer:

Pavlova-9 [17]4 years ago

3 0

5 because if you do math then you get that

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roman has 2 red pencils in his backpack.if this is 25%of the pencils ,how meny pencils are in his backpack

Lesechka [4]

There are 8 pencils in his backpack

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3 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!

4 0

3 years ago

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Evaluate 3b-2a-9 a= 1/2 and b=7

ale4655 [162]

Answer

Step-by-step explanation:

I hope this helps.

7 0

3 years ago

Solve the equation for P: 3k=7Q+6p

iren [92.7K]

3k=7Q+6p

Solve for P which means we have to make 'p' alone

$3k=7Q+6p\\\\ Add 7Q on both sides \\\\ 3k - 7Q = 7Q - 7Q + 6p\\\\ \left ( 3k - 7Q \right ) = 6p\\\\ Divide both side by 6 to make p alone \\\\ \frac{\left ( 3k - 7Q \right )}{6} = \frac{6p}{6}\\\\ p = \frac{\left ( 3k - 7Q \right )}{6}$

7 0

3 years ago

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PLEASE HELP! MATH

nikdorinn [45]

6x^4+9x^2+12x/3x = 2x^3+3x+4

So the answer is D. 2x3 + 3x + 4

5 0

3 years ago

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