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Whitepunk [10]
3 years ago
13

The point A(-6, 8) has been transformed using the composition r(90,O) counterclockwise ∘Rx−axis . Where is A'?

Mathematics
1 answer:
Vikki [24]3 years ago
7 0
A would now be a positive 6
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3, 1-을<br>6. +<br>9. 1-(급+음<br>12. 37 -27​
umka21 [38]

Answer:

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6 0
3 years ago
Write the quadratic equation whose roots are 1 and -5, and whose leading coefficient is 3.
Anna35 [415]

Step-by-step explanation:

they're is a neat little trick.

but first of all, "roots" means the 0 solutions of the equation. like

ax² + bx + c = 0

there are 2 solutions for x to make the whole expression equal 0.

these are the "roots".

now to the little trick :

when is a multiplication resulting in 0 ?

when at least one of the factors is 0.

and any quadratic expression can be written as multiplication of 2 factors. like

c(x - a)(x - b) = cx² - cax - cbx + cab

what are the "roots" or 0 points ?

either

c = 0

x - a = 0 | x = a

x - b = 0 | x = b

the leading coefficient = 3.

that means nothing else than c = 3.

root = 1 means (x - 1) is one factor.

root = -5 means (x + 5) is the other factor.

so, we have

3(x - 1)(x + 5) = 3x² + 12x - 15

and the equation is

3x² + 12x - 15 = 0

or

3x² + 12x = 15

7 0
2 years ago
Which represents the area of the figure?
Ira Lisetskai [31]

Answer:

A

Step-by-step explanation:

The correct option is (A).

5 0
2 years ago
Read 2 more answers
The highest common factor (HCF) of 140 and x is 20
Zielflug [23.3K]

Answer:

x=60

Step-by-step explanation:

Given that ,

HCF of 140 and x=20,

LCM of 140 and x =420.

Solution:

It is known that the product of two numbers is equal to the product of their HCF and LCM, i-e:

140*x= HCF*LCM

Putting in the values provided,

140*x=20*420

140*x=8400

x=8400/140

x=60

Hence the variable x is 60.

5 0
3 years ago
Rewrite the expression 4+<img src="https://tex.z-dn.net/?f=%5Csqrt%7B16-%284%29%285%29%7D" id="TexFormula1" title="\sqrt{16-(4)(
Inessa05 [86]

Answer:

2+i

Step-by-step explanation:

Given the expression:

\dfrac{4+\sqrt{16-(4)(5)}}{2}

To find:

The expression of above complex number in standard form a+bi.

Solution:

First of all, learn the concept of i (pronounced as <em>iota</em>) which is used to represent the complex numbers. Especially the imaginary part of the complex number is represented by i.

Value of i =\sqrt{-1}.

Now, let us consider the given expression:

\dfrac{4+\sqrt{16-(4)(5)}}{2}\\\Rightarrow \dfrac{4+\sqrt{16-(4\times 5)}}{2}\\\Rightarrow \dfrac{4+\sqrt{16-20}}{2}\\\Rightarrow \dfrac{4+\sqrt{-4}}{2}\\\Rightarrow \dfrac{4+\sqrt{(-1)(4)}}{2}\\\Rightarrow \dfrac{4+\sqrt{(-1)}\sqrt4}{2}\\\Rightarrow \dfrac{4+\sqrt4i}{2} \ \ \ \ \ (\because \sqrt{-1} =i) \\\Rightarrow \dfrac{4+2i}{2}\\\Rightarrow 2+i

So, the given expression in standard form is 2+i.

Let us compare with standard form a+bi so we get a =2, b =1.

\therefore The standard form of

\dfrac{4+\sqrt{16-(4)(5)}}{2}

is: \bold{2+i}

8 0
3 years ago
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