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3 years ago

Plis only answer correctly

Mathematics

2 answers:

gregori [183]3 years ago

8 0

Answer:

Step-by-step explanation:

dkimiunuhbubhwswdqwduoewdWEC W vcuuhdcosv

Gemiola [76]3 years ago

3 0

The answer is B
Because it’s the most likely

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A dilation with a scale factor of 3. 5 and centered at the origin is applied to PQ¯¯¯¯¯ with endpoints P(2, 5) and Q(2, 1)

velikii [3]

Answer: The answer is P'(7, 17.5) and Q'(7, 3.5).

Step-by-step explanation: Given that a line segment PQ is dilated with a scale factor of 3.5 where origin is the centre of dilation.

The end points of segment PQ are P(2, 5) and Q(2, 1).

Therefore, after dilation, the coordinates of the end points become

Thus, the coordinates of P' are (7, 17.5) and the co-ordinates of Q' are (7, 3.5).

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2 years ago

WILL GIVE BRAINLESS <br>factor the polynomial

Svetach [21]

Answer:

Step-by-step explanation:

This is factor by grouping. In factor by grouping, write a quadratic trinomil as 4 terms and group by parenthesis. Then factor by GCF in each pair. If the two parenthesis match, the factoring has worked and the factors will be the GCFs as one and the parenthesis as one other.

The factors here are the GCFs $2x^2$ and -3 as $(2x^2-3)$ and the parenthesis (x-5).

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3 years ago

How does charge Build Up?

Masteriza [31]

Answer: Charges build up when negative electrons are transferred from one object to another. The object that gives up electrons becomes positively charged, and the object that accepts the electrons becomes negatively charged

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3 years ago

Solve the following equations by factorisation method.Only factorisation not dharacharya.

Luba_88 [7]

Hello, please consider the following.

When $x_1$ and $x_2$ are two roots, we can factorise as

$ax^2+bx+c=a(x-x_1)(x-x_2)=a(x^2-(x_1+x_2)x+x_1x_2)=0$

So for the first equation, we can say that the sum of the zeros is

$\dfrac{a^2+b^2}{2}=\dfrac{a^2}{2}+\dfrac{b^2}{2}$

and the product is

$\dfrac{a^2b^2}{4}=\dfrac{a^2}{2}\dfrac{b^2}{2}$

So we can factorise as below.

$4x^2-2(a^2+b^2)x+a^2b^2=(2x-a^2)(2x-b^2)=0$

And the solutions are

$\boxed{\sf \n\bf \ \dfrac{a^2}{2} \ \ and \ \ \dfrac{b^2}{2}}$

For the second equation, we will complete the square and put the constant on the right side and take the root.

Let's do it!

$9x^2-9(a+b)x+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2-9\dfrac{(a+b)^2}{4}+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9(a+b)^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9a^2+18ab+9b^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{a^2+b^2-2ab}{4}=\dfrac{(a-b)^2}{4}\\\\(x-\dfrac{a+b}{2})^2=\dfrac{(a-b)^2}{2^2\cdot 3^2}$