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3 years ago

Heyyy I need helppp plzzz

Mathematics

2 answers:

finlep [7]3 years ago

5 0

x + 2x = 109

mark brainliest

Alecsey [184]3 years ago

4 0

Answer:

x + 2x = 109

Step-by-step explanation:

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Answer: The answer would be $556.

Step-by-step explanation:

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3 years ago

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iVinArrow [24]

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3 years ago

Which figure is the image produced by applying the composition T 0,3 o R0,90 to figure R?

Lyrx [107]

We are given original image R.

It is being translated by (0,3) first and then rotated by a positive angle 90 degrees.

Translation by (0,3) represents (x,y) --> (x, y+3) rule.

Positive 90 degree rotation represents, counterclockwise rotation.

The rule for counterclockwise rotation is (x,y) --> (-y,x).

Therefore, final rule for would be

(x,y) --> ( -y,x+3 )

Let us take a coordinate of R on y-axis as (0,-4).

Now if we apply rule (x,y) --> ( -y,x+3 ) we get

(0,-4) --> (-(-4), 0+3) = (4,3).

Let us check the figure with coordinate (4,3).

We can clearly see that Figure H has transformed coordinate (4,3).

<h3>Therefore, correct option is first option A. figure H.</h3>

8 0

3 years ago

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MakcuM [25]

Answer:

6562331215

Step-by-step explanation:

3 0

3 years ago

What does the fundamental theorem of algebra illustrate?

UNO [17]

Answer:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

Step-by-step explanation:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

We have to find the roots of this given equation.

If a quadratic equation is of the form $ax^{2}+bx +c=0$

Its roots are $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Here the given equation is $2x^{2}-4x-1$ = 0

a = 2

b = -4

c = -1

If the roots are $x_{1} and x_{2}$ , then

$x_{1}$ = $\frac{-2+\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}$

= $\frac{4 +\sqrt{24}}{4}$

= $\frac{2+\sqrt{6} }{2}$

$x_{2}$ = $\frac{-2-\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}$

= $\frac{4 +\sqrt{8}}{4}$

= $\frac{2-\sqrt{6} }{2}$

These are the two roots of the equation.

6 0

3 years ago