The points (1, 3) and (−2, 6) lie on a line.where does the line cross the x-axis

Identify a reflection about the x-axis or a reflection about the y-axis of f(x) = √-x by observing the equation of the function

olga55 [171]

Answer:

B) The base graph has been reflected about the y-axis

Step-by-step explanation:

We are given the function, $f(x)=\sqrt{-x}$ .

Now, as we know,

The new function after transformation is $g(x)=f(-x)=\sqrt{x}$ .

<em>As, the function f(x) is changing to g(x) = f(-x)</em> and from the graph below, we see that,

The base function is reflected across y-axis.

Hence, option B is correct.

4 0

4 years ago

Which linear equation below represents the line that goes through the point (3.4) and has a slope of 2?

Ad libitum [116K]

Answer:

Linear equation with a slope of 2 that goes through the point (3, 4) is $y = 2\cdot x -2$ .

Step-by-step explanation:

From statement we know the slope of the line and a point contained in it. Using the slope-point equation of the line is the quickest approach to determine the appropriate equation, whose expression is:

$y-y_{o} = m \cdot (x-x_{o})$

Where:

$m$ - Slope, dimensionless.

$x_{o}$ , $y_{o}$ - Components of given point, dimensionless.

$x$ , $y$ - Independent and dependent variable, dimensionless.

If we know that $m = 2$ , $x_{o} = 3$ and $y_{o} = 4$ , the linear equation is found after algebraic handling:

1) $y-4 = 2\cdot (x-3)$ Given

2) $y = 2\cdot (x-3) +4$ Compatibility with Addition/Existence of Additive Inverse/Modulative Property

3) $y = 2\cdot x -2$ Distributive Property/ $(-a)\cdot b = -a\cdot b$ /Definition of sum/Result

Linear equation with a slope of 2 that goes through the point (3, 4) is $y = 2\cdot x -2$ .

7 0

4 years ago

A basket contains five apples and seven peaches. Four of the apples and two of the peaches are rotten. You randomly pick a piece

mash [69]

2/35possible outcomes:5*7=35fresh or apple = 22/35

6 0

4 years ago

Read 2 more answers

9. Consider the line y = 4x+9. If a second line is perpendicular to this one, what is its slope? please show your work, i will m

zlopas [31]

Answer:

Slope of the line perpendicular to the given line = $- \frac{1}{4}$

Step-by-step explanation:

If two are lines are perpendicular to each other,

the product of their slopes = - 1 .

That is ,

$m_ 1 \times m_2 = -1$

Slope of the given line :

$m_ 1 = 4$

Hence slope of the line perpendicular to it :

$4 \times m_ 2 = - 1 \\\\\frac{4}{4} \times m_2 = - \frac{1}{4} \\\\1 \times m_2 = - \frac{1}{4}\\\\m_ 2 = \frac{-1}{4}$

6 0

3 years ago

Read 2 more answers

The function h(t) = -16t^2 + 16t represents the height (in feet) of a horse (t) seconds after it jumps during a steeplechase.

Vladimir79 [104]

A.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.

h'(t) = -32t + 16

When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.

-32t + 16 = 0

-32t = -16

t = 0.5 seconds

b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.

h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet

If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.

c.) We know that the horse is in the air whenever h(t) is greater than 0.

-16t^2 + 16t = 0

-16t(t-1)=0

t = 0 and 1

So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.

7 0

3 years ago