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Mathematics

1 answer:

zheka24 [161]2 years ago

6 0

Answer:

Step-by-step explanation:

$3x+7\\\\3\left(x+\dfrac{7}{3}\right)=(3)(x)+(\not3)\left(\dfrac{7}{\not3}\right)=3x+7\qquad\boxed{YES}\\_{\text{used the distributive property: a(b + c) = ab + ac}}\\\\x+x+x+5+1=(x+x+x)+(5+1)=3x+6\neq3x+7\qquad\boxed{NO}\\\\8x+3-5x+4=(8x-5x)+(3+4)=3x+7\qquad\boxed{YES}\\\\7x+3\neq3x+7\qquad\boxed{NO}$

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Tran is solving the quadratic equation 2x2 – 4x – 3 = 0 by completing the square. His first four steps are shown in the table.

Nady [450]

The following steps of solving for the roots of 2x² - 4x -3 = 0 were retrieved from another source Step 1 2x² - 4x = 3 Step 2 2(x² - 2x) = 3 Step 3 2(x² - 2x + 1) = 3 + 1 Step 4 2(x - 1)² = 4 From this, we can see that on Step 3, Tran made a mistake of adding 1 to 3. As we can see, 2(x² - 2x + 1) = 2x² - 4x + 2. That means, instead of adding 1, it should have been 2. Therefore, the step that Tran first made an error is Step 3<span>.</span>

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3 years ago

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The black graph is the graph of y=f(x). Choose the equation for the red graph.<br> Please help me!!!

dusya [7]

Answer:

Step-by-step explanation:

The red graph is a horizontal translation of 5 units left followed by a reflection in the x- axis.

Given f(x) then f(x + a) is a horizontal translation of f(x)

• If a > 0 then a shift to the left of a units

• If a < 0 then a shift to the right of a units

The black graph is shifted 5 units left, thus

f(x) → f(x + 5)

Under a reflection in the x- axis

a point (x, y ) → (x, - y )

Note that the y- coordinate of the image is the negative of the original

Note also that $\frac{y}{-1}$ = - y, thus

$\frac{y}{-1}$ = f(x + 5) → B

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3 years ago

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Which equation represents the equation of the parabola with focus (-3 3) and directrix y=7?

Artemon [7]

Answer:

The equation $y=\frac{-x^2-6x+31}{8}$ represents the equation of the parabola with focus (-3, 3) and directrix y = 7.

Step-by-step explanation:

To find the equation of the parabola with focus (-3, 3) and directrix y = 7. We start by assuming a general point on the parabola (x, y).

Using the distance formula $d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }$ , we find that the distance between (x, y) is

$\sqrt{(x+3)^2+(y-3)^2}$

and the distance between (x, y) and the directrix y = 7 is

$\sqrt{(y-7)^2}$ .

On the parabola, these distances are equal so, we solve for y:

$\sqrt{(x+3)^2+(y-3)^2}=\sqrt{(y-7)^2}\\\\\left(\sqrt{\left(x+3\right)^2+\left(y-3\right)^2}\right)^2=\left(\sqrt{\left(y-7\right)^2}\right)^2\\\\x^2+6x+y^2+18-6y=\left(y-7\right)^2\\\\x^2+6x+y^2+18-6y=y^2-14y+49\\\\y=\frac{-x^2-6x+31}{8}$