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3 years ago

Write your answer for y.

Mathematics

1 answer:

Nonamiya [84]3 years ago

4 0

Answer:

The coordinates of Y' will be: A'(4, -3)

Step-by-step explanation:

Triangle XYZ with vertices

X(-5, -2)

Y(-3, 4)

Z(-1, 1)

We have to determine the answer for the image Y' of the point Y(-3, 4) with respect to the line mirror y=x.

We know that when P(x, y) is reflected in y = x, we get P'(y, x)

i.e.

The rule of y=x:

P(x, y) → P'(y, x)

As we are given that Y(-3, 4), so the coordinates of Y' will be:

A(-3, 4) → A'(4, -3)

Therefore, the coordinates of Y' will be: A'(4, -3)

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4x-5=7+4y How do you solve?

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solve for x.

4x−5=7+4y

Add 5 to both sides.

4x−5+5=4y+7+5

4x=4y+12

Divide both sides by 4.

4x/4=4y+12/4

x=y+3

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At work jenine helps 6 customers every 40 minutes. How many customers can she expect to help in 240 minutes before her break?

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Step-by-step explanation:

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Evalute costheta if sintheta = (sqrt5)/3

torisob [31]

Answer:

$\large\boxed{\cos\theta=\pm\dfrac{2}{3}}$

Step-by-step explanation:

Use $\sin^2x+\cos^2x=1$ .

We have

$\sin\theta=\dfrac{\sqrt5}{3}$

Substitute:

$\left(\dfrac{\sqrt5}{3}\right)^2+\cos^2\theta=1\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(\sqrt5)^2}{3^2}+\cos^2\theta=1\qquad\text{use}\ (\sqrt{a})^2=a\\\\\dfrac{5}{9}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{5}{9}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{9}{9}-\dfrac{5}{9}\\\\\cos^2\theta=\dfrac{4}{9}\to \cos\theta=\pm\sqrt{\dfrac{4}{9}}\\\\\cos\theta=\pm\dfrac{\sqrt4}{\sqrt9}\\\\\cos\theta=\pm\dfrac{2}{3}$

3 0

3 years ago

Read 2 more answers

Identify the interval on which the quadratic function is positive.

Alenkasestr [34]

Answer:

$\textsf{1. \quad Solution: $1 < x < 4$,\quad Interval notation: $(1, 4)$}$

$\textsf{2. \quad Solution: $-2 < x < 4$,\quad Interval notation: $(-2, 4)$}$

Step-by-step explanation:

<h3>Question 1</h3>

The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies -7x^2+35x-28& = 0\\-7(x^2-5x+4)& = 0\\x^2-5x+4& = 0\\x^2-x-4x+4& = 0\\x(x-1)-4(x-1)&= 0\\(x-1)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x-1=0 \implies x=1$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is positive is:

Solution: 1 < x < 4
Interval notation: (1, 4)

<h3>Question 2</h3>

The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.

The zeros of the quadratic function are the points at which the parabola crosses the x-axis.

As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.

To find the zeros of the given quadratic function, substitute y = 0 and factor:

$\begin{aligned}y&= 0\\\implies 2x^2-4x-16& = 0\\2(x^2-2x-8)& = 0\\x^2-2x-8& = 0\\x^2-4x+2-8& = 0\\x(x-4)+2(x-4)&= 0\\(x+2)(x-4)& = 0\end{aligned}$

Apply the zero-product property:

$\implies x+2=0 \implies x=-2$

$\implies x-4=0 \implies x=4$

Therefore, the interval on which the function is negative is:

Solution: -2 < x < 4
Interval notation: (-2, 4)

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1 year ago

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scoundrel [369]

Answer:

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Step-by-step explanation:

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